Calculus of Variations: Part II

by greenturtle3141, Jul 31, 2022, 9:11 PM

Calculus of Variations Series

Part I: https://artofproblemsolving.com/community/c2532359h2742841
Part II: This

Reading Level: 6/5

Prerequisites:

You must be familiar with the following:

Basic topology (open/closed sets)
liminf and limsup
$L^p$ space

It is highly recommended that you are familiar with:

Bolzano-Weierstrass; sequential compactness
Fatou's Lemma
Holder's Inequality

The following would be nice to be familiar with, but we're going to review them regardless:

Convex functions
Riesz Representation Theorem for $L^p$ spaces
Geometric Hahn-Banach theorem

Part 1: The Direct Method

Last time, we left off with the following issue: Using the Euler-Lagrange equation, we maanged to find pretty convincing candidates for the minimizer of integral functionals such as $F(u) = \int_0^1 (u(x)+u'(x))^2\,dx$ . However, who is to say that the candidate is actually a minimizer?

Typically, the process for actually finding a minimizer for, say, a function like $f(x) = x^4+x$ , goes as follows:

Prove that a minimum actually exists.
Using a calculus argument, obtain a list of possible candidates for the minimizer.
Pick the candidate that produces the minimum value of all the candidates, and conclude by (1) that it produces the minimum value, and hence is a minimizer.

In the case of $x^4+x$ , we can do (1) by using some kind of extreme value argument. Step (2) is done by solving $f'(x) = 0$ . (3) is done simply.

In the case of a function(al) in infinite dimensions like $\int_0^1 (u(x)+u'(x))^2\,dx$ , we're kinda in trouble. Step (2) is handled by the Euler-Lagrange equation, and (3) is a trivial process. But how do we go about (1)?

The Direct Method of the Calculus of Variations handles this. I'm going to state it just so you know what's coming, so don't run away if you have no idea what any of it means:

Obtain a lower bound on the functional $F:X \to \overline{\mathbb{R}}$ .
Deduce that $F$ has an infimum, and thus admits a minimizing sequence $u_n$ such that $F(u_n) \to \inf\{F(u) : u \in X\}$ .
Find a notion of convergence $\tau$ such that we get the following compactness result: If $\{u_n\}_n$ is a sequence such that $\{F(u_n)\}_n$ is a bounded sequence, then there exists a subsequence $u_{n_k}$ such that $u_{n_k} \stackrel\tau\to u$ for some $u \in X$ . Apply this compactness result to the sequence $\{u_n\}_n$ from the previous step.
Prove that $F$ is sequentially lower semi-continuous with respect to $\tau$ .
We're done!

Let's translate this procedure to English.

This is exactly what it sounds like: Find some really small $c$ such that $F(x) \geq c$ for all $x \in X$ .
Any set that is bounded from below has an infimum, so there must exist the infimum $m = \inf_{x \in X} F(x)$ . Of course, this does not necessarily mean that infimum can be obtained... after all, that's precisely what we're trying to prove! What this does imply is that for all $n \in \mathbb{N}$ , I can find some $x_n \in X$ for which $m \leq F(x_n) < m + \frac1n$ . In particular, this means that $F(x_n) \to m$ , and we call $\{x_n\}_{n=1}^\infty$ a minimizing sequence.
Let's shelve the motivation for compactness for now. Once we have a minimizing sequence, the idea is to prove that $F(x_n)$ is sequentially lower-semicontinuous, so that in theory we can turn the $F(x_n) \to m$ into $F(x) = m$ for some $x$ .

As a reminder, a function $f$ is sequentially lower-semicontinuous at some $x_0$ if, for every $y_n \to y_0$ , we have that $\liminf_{n \to \infty} f(y_n) \geq f(y_0)$ .

This idea doesn't make too much sense yet though, for two reasons. The first problem is that the minimizing sequence we have, $x_n$ , doesn't necessarily converge to some $x_0$ . The point of step 3 of the procedure is to fix this! In this step, we show that there is a subsequence $x_{n_k}$ that does converge to some $x_0$ (and in theory this should be a minimizer of $F$ ). This is what it means to prove a compactness result: Given that $F(x_n)$ is bounded, prove that there is a convergent subsequence of $x_n$ .

But there is a second issue. We keep talking about "convergence of $x_{n_k} \to x_0$ ", both in regards to the compactness result and the "sequential lower-semicontinuity". What kind of convergence is this? All this discussion is moot if I don't specify what notion convergence this is.

If we're working in $\mathbb{R}$ , then there isn't much of a choice of convergence besides the standard one. But for function spaces like $L^1(0,1)$ , you have a number of choices, like convergence in measure and convergence in $L^p(0,1)$ . Some of these choices for convergence are stronger than others. The weaker the convergence you choose is, the easier it will be to prove the compactness result. So we choose some convergence $\tau$ that is weak enough for $x_n$ to admit a convergence subsequence $x_{n_k} \to x_0$ , provided that $F(x_n)$ is bounded.

Why the greek letter tau?
By "choosing a convergence", what's really going on behind the scenes is that I'm choosing a topology.

(Doesn't make 100% sense because almost-everywhere convergence does not induce a topology, but whatever.)
Now the final nail in the coffin is to prove that $F$ is sequentially lower-semicontinuous with respect to the convergence $\tau$ that we chose. That is, if $y_n \stackrel\tau\to y_0$ , then $\liminf_{n \to \infty} F(y_n) \geq y_0$ . The stronger the convergence $\tau$ is, the easier it is to prove sequential lower-semicontinuity. This means that there is a delicate balance in choosing $\tau$ : It cannot be too weak nor too strong.
Now we're obviously done! ... I'm kidding, it's not that obvious. Here's why.
- We have a minimizing sequence $x_n$ with $F(x_n) \to m$ .
- By boundedness of $F(x_n)$ , we may apply the compactness result proved in Step 3 to deduce that there exists a subsequence $x_{n_k}$ such that $x_{n_k} \stackrel\tau\to x_0$ for some $x_0 \in X$ .
- By sequential lower-semicontinuity of $F$ with respect to $\tau$ , we now get that:
  $\liminf_{k \to \infty} F(x_{n_k}) \geq F(x_0)$
- One one hand, $F(x_0) \geq m$ since $m$ is an infimum (in particular it is a lower bound). On the other hand, $\liminf_{k \to \infty} F(x_{n_k}) = m$ , because $\lim_{n \to \infty} F(x_n) = m$ (extracting a subsequence doesn't change the limit). Thus $m \geq F(x_0) \geq m$ , meaning that $F(x_0) = m$ . So indeed there exists $x_0 \in X$ for which $F(x_0)$ is the global minimum of $F$ , i.e. the minimum is obtained, i.e. $F$ has a minimum. This is precisely what we needed to prove!

A Stupid Example

Let our space be $X = \mathbb{R}^2$ , and let $f(x,y) = x^2+y^2$ . We claim that $f$ has a minimum. Shocking, I know.

Our recipe above will be overkill for this example, but let's just do it so you're more familiar with how this flows.

Is $f$ bounded from below? Yep, because $x^2 \geq 0$ and $y^2 \geq 0$ , so $f(x,y) \geq 0$ for all $x,y$ .
Blah blah minimizing sequence blah, we don't need to do any work here, it's all automatic.
Now we need a compactness result of the following form: If $\{(x_n,y_n)\}_{n=1}^\infty \in \mathbb{R}^2$ is a sequence for which $f(x_n,y_n)$ is bounded (which we may write as "\sup_{n \in \mathbb{N}} f(x_n,y_n) < \infty" to sound like the cool kids), then we may find a subsequence $(x_{n_k},y_{n_k})$ for which $(x_{n_k},y_{n_k}) \stackrel\tau\to (x_0,y_0) \in \mathbb{R}^2$ .

What convergence $\tau$ should we choose? ...well, we really don't have much of a choice here, do we? Let's just pick the standard Euclidean metric/norm $\|\cdot\|_2$ , i.e. the most obvious choice for convergence in $\mathbb{R}^2$ that there is.

Alright, now suppose $\sup_{n \in \mathbb{N}} f(x_n,y_n) = M < \infty$ . Then that means that $x_n^2 + y_n^2 \leq M$ for all $n$ , so $(x_n,y_n) \in B((0,0), \sqrt{M})$ for all $n$ . In particular the sequence $(x_n,y_n)$ must be bounded, hence by Bolzano-Weierstrass we may find a converging subsequence $(x_{n_k},y_{n_k}) \to (x_0,y_0) \in \mathbb{R}^2$ . Tada!
Lastly we need sequential lower-semicontinuity of $x^2+y^2$ . But uh, I mean, it's obviously continuous with respect to the standard Euclidean notion of convergence, so it's certainly lower-semicontinuous and hence sequentially lower-semicontinuous.
We're done, a minimum exists! We don't need to do work here either.

As you can see, the work we need to do ourselves boils down to (1) finding a lower bound for $F$ , (2) proving a compactness result with respect to an appropriate notion of convergence $\tau$ , and (3) proving lower-semicontinuity with respect to $\tau$ .

A Not So Stupid Example...?

Problem 1: Let $\Omega$ be an open and bounded interval, and let $F:L^2(\Omega) \to \mathbb{R}$ be an integral functional defined as:
$F(u) := \int_\Omega u(x)^2\,dx$ Prove that $F$ has a minimum.

(From now on, we will abuse notation by dropping the $(x)$ when the meaning is clear, so you'll see stuff like $\int_\Omega u^2\,dx$ .)

Well ok, let's try applying the Direct Method (even though, again, it is very unnecessary...).

Lower Bound

Since $u^2 \geq 0$ for all $u$ , it is clear that $F(u) \geq 0$ for all $u \in L^2(\Omega)$ , hence $0$ is a lower bound.

Compactness

Suppose $\sup_{n \in \mathbb{N}} F(u_n) = M < \infty$ . Then $\int_\Omega u_n^2\,dx \leq M$ for all $n$ . In particular $\{u_n\}$ is bounded in $L^2(\Omega)$ . Now we just need to find a subsequence that... uh... er... eh? What convergence do we even choose, and how in the world do we find a converging subsequence? The last example was easy because of Bolzano-Weierstrass, but $L^2(\Omega)$ is an infinite-dimensional normed space! What in the world could we possibly do???

Sequential Lower-Semicontinuity

Er... maybe we can try and take a stab at this. Let's choose $\tau$ to be "convergence in $L^2(\Omega)$ " and see if it works.

Suppose $u_n \to u$ in $L^2(\Omega)$ . Let $L = \liminf_{n \to \infty} \int_\Omega u_n^2\,dx$ . We want to prove that $L \geq \int_\Omega u^2\,dx$ .

By properties of the liminf, there is a subsequence $u_{n_k}$ for which $L = \lim_{k \to \infty} \int_\Omega u_{n_k}^2\,dx$ . We still have $u_{n_k} \stackrel{L^2(\Omega)}\to u$ , so by some random theorem ("converse of dominated convergence"), there exists a further subsequence $u_{n_{k_l}}$ that converges almost-everywhere to $u$ . Now by Fatou's Lemma:
$L = \lim_{l \to \infty} \int_\Omega u_{n_{k_l}}^2\,dx \geq \int_\Omega \liminf_{l \to \infty} u_{n_{k_l}}^2\,dx = \int_\Omega u^2\,dx$ Half yay?

...

Well that didn't go very well. The problem is that infinite-dimensional calculus is not easy (wow what a surprise).

By choosing $\tau$ to be $L^2(\Omega)$ convergence, we managed to prove the sequential lower-semicontinuity, but we were unable to show compactness. In fact, this convergence is actually too strong to get a compactness result! Finding a counterexample is left as an exercise.

What about convergence in measure?

So, we need to choose some convergence weaker than $L^2(\Omega)$ ... and I have just the thing!

Part 2: Weak $L^p$ Convergence and Weak Compactness

I'm going to try and motivate where the heck this comes from. This motivation is not essential, so feel free to skip straight to the definition of weak convergence if this flies over your head.

One of the biggest secrets of math is that being an object is closely tied to how you are related to other objects.

As an example in linear algebra, consider a point $\vec{x} = (x_1,x_2,\cdots,x_N) \in \mathbb{R}^N$ . By giving you this point $\vec{x}$ , you... know what $\vec{x}$ . I mean, duh. But what if I only told you how $\vec{x}$ is related to other things? Specifically, what if for every element $T$ in the dual space of $\mathbb{R}^N$ (i.e. the space of all linear maps $\mathbb{R}^N \to \mathbb{R}$ ), I told you the result of $T\vec{x}$ ? Can you deduce $\vec{x}$ ? Of course! If we take the map $T = e_k^*$ (the linear map that just spits out the $k$ th coordinate) for each $1 \leq k \leq N$ , then this tells us all the components of $\vec{x}$ , hence the information we probe about $\vec{x}$ by attacking it with dual-space elements tells us everything about $\vec{x}$ .

There is an analogue of this for $L^p$ space. If $1 < p < \infty$ and $q$ is the Holder conjugate of $p$ (so that $1/p+1/q = 1$ ), then the dual space of $L^p(E)$ is the space of linear functions that take the form $Tu = \int_E uv\,dx$ , where $v \in L^q(E)$ . By Holder's inequality, $Tu$ is always real (i.e. the integral converges in $\mathbb{R}$ ) because $\int_E uv\,dx \leq \|u\|_{L^p(E)}\|v\|_{L^q(E)} < \infty$ , hence why we use the Holder conjugate. Now, something called the Riesz Representation Theorem essentially says that being a function $u \in L^q(E)$ is essentially the same as being a linear map $v \mapsto \int_E uv\,dx$ . What this implies is that a function $u \in L^p(E)$ is uniquely determined by all the information by attacking $u$ with an element $v \in L^q(E)$ via $\int_E uv\,dx$ .

We're going to need Riesz-Representation later, so I'm going to state it formally here.

Theorem (RRT):

For every $v \in L^q(E)$ , the map $T_v: u \mapsto \int_E uv\,dx$ is a linear map $L^p(E) \to \mathbb{R}$ (duh), and is continuous (why?).
Conversely (!!!), for every continuous linear map $T:L^p(E) \to \mathbb{R}$ , there is a unique $v \in L^q(E)$ for which $T = T_v$ .

Continuity?

Anyways, the motto here is that looking at $u \in L^p(E)$ is closely tied with looking at $\int_E uv\,dx$ for each $v \in L^q(E)$ . And hopefully this is enough to motivate the star of the show...

Definition (Weak $L^p$ convergence): Let $E \subseteq \mathbb{R}^N$ be measurable, let $1 \leq p < \infty$ , and consider a sequence $u_n \in L^p(E)$ . We say that $u_n$ converges weakly in $L^p(E)$ to some function $u \in L^p(E)$ if
$\lim_{n \to \infty} \int_E u_nv\,dx = \int_E uv\,dx$ for all $v \in L^q(E)$ , where $q$ is the Holder conjugate of $p$ . This is equivalent to
$\lim_{n \to \infty} \int_E (u_n-u)v\,dx = 0 \qquad \forall v \in L^q(E).$ If $u_n$ converges weakly to $u$ in $L^p(E)$ , we write " $u_n \rightharpoonup u$ in $L^p(E)$ ".

(You can also define this in the same way for $p=+\infty$ but it's not as wholesome so we write $\stackrel{\ast}\rightharpoonup$ instead. We won't be needing this.)

Well ok, interesting. But before we accept $\rightharpoonup$ as a member of the "notion of convergence family", we gotta make sure that it's a sane notion of convergence. You know how there are some weird topological spaces in which limits aren't unique? Let's make sure that doesn't happen.

Proposition: If $u_n \rightharpoonup u$ and $u_n \rightharpoonup \bar{u}$ , then $u = \bar{u}$ almost everywhere.

Proof.

Fix $v \in L^q(E)$ . Then $\lim_{n \to \infty} \int_E (u_n-u)v\,dx = 0$ and $\lim_{n \to \infty} \int_E (\bar{u}-u_n)v\,dx = 0$ . By adding these together, we get that $\lim_{n \to \infty} \int_E (\bar{u}-u)v\,dx = 0$ . But there's no $n$ in this equation, so this is just saying that $\int_E (\bar{u}-u)v\,dx = 0$ , and this holds for all $v \in L^q(E)$ .

We can conclude by Riesz-Representation. Alternatively... $\square$

Alright cool. Great. But if I'm calling this weird thing weak $L^p$ convergence, then it better be literally weaker than $L^p$ convergence (which we sometimes call "strong" $L^p$ convergence). Let's check that.

Proposition: If $u_n \to u$ in $L^p(E)$ , then $u_n \rightharpoonup u$ in $L^p(E)$ .

Proof. Fix an arbitrary $v \in L^q(E)$ . Then by Holder:
$\int_E (u_n-u)v\,dx \leq \|u_n-u\|_{L^p(E)}\|v\|_{L^q(E)}$ But $\|u_n-u\|_{L^p(E)}\|v\|_{L^q(E)} \to 0$ because $\|u_n-u\|_{L^p(E)} \to 0$ and $\|v\|_{L^q(E)} < \infty$ by virtue of $v \in L^q(E)$ . $\square$

Wonderful! But why do we care about weak convergence? Remember, we need a sufficiently weak notion of convergence to get a sort of compactness result. Indeed, we get it!

Theorem (Weak Compactness in $L^p$ ): Let $u_n \in L^p(E)$ be a sequence for which $\sup_{n \in \mathbb{N}}\|u_n\|_{L^p(E)} = M < \infty$ . Then there exists a subsequence $u_{n_k}$ for which $u_{n_k} \rightharpoonup u$ in $L^p(E)$ .

Proof.

This is a tough proof. Two key ideas:

Infinite-dimensional spacess are hard. But $L^q(E)$ is separable! This means that there exists a sequence $v_m \in L^q(E)$ that is dense in $L^q(E)$ . Countable is good! We like countable. There is hope.
Instead of dealing with functions, what if we played with linear maps instead? Riesz Representation says that it's basically the same thing, and moreover linear maps output real numbers, which we know a lot about. (This is what weakening the convergence allows us to exploit!!!)

Alright so first, instead of dealing with $u_n$ , let's deal with the linear map $T_n: v \mapsto \int_E u_nv\,dx$ . The sketch here is that we're going to find a subsequence of $T_n$ that converges at enough points to some $T$ . Then by Riesz-Representation, we'll switch back from $T$ to $u$ .

By enough points, I just mean the dense subset $v_m$ ! Let's do this step-by-step.

First handle $v_1$ . $T_n(v_1) = \int_E u_nv_1\,dx \leq \|u_n\|_{L^p(E)}\|v_1\|_{L^q(E)} \leq M\|v_1\|_{L^q(E)} < \infty$ , so the sequence of real numbers $\{T_n(v_1)\}_{n=1}^\infty$ is bounded! By Bolzano-Weierstrass, we may now find a subsequence $T_{1,n}$ of $T_n$ such that $T_{1,n}(v_1) \to L_1 \in \mathbb{R}$ .
Next up is $v_2$ . By the same argument, $T_{1,n}(v_2)$ is a bounded sequence, so we may find a subsequence $T_{2,n}$ of $T_{1,n}$ such that $T_{2,n}(v_2) \to L_2 \in \mathbb{R}$ . Note that since subsequence extraction preserves limits, we still have $T_{2,n}(v_1) \to L_1$ .
Next is $v_3$ . By the same argument, we find a subsequence $T_{3,n}$ of $T_{2,n}$ such that $T_{3,n}(v_3) = L_3$ .
etc. etc. etc., do this forever.

Hm, doing this forever isn't really helpful. Since this process never ends, I don't end up with a good sequence... or do I? Analysis aficionados know what's coming next: A classic diagonalization argument! Ultimately we now take the subsequence $T_{n_k} := T_{k,k}$ . Convince yourself that $T_{n_k}(v_i) \to L_i$ for all $i$ !

Now we want to define the "limiting linear map" $T:L^q(E) \to \mathbb{R}$ . Our natural guess is:
$T(v) := \lim_{k \to \infty} T_{n_k}(v)$ Does this work? Well, for all $i$ , we get that $T(v_i) = \lim_{k \to \infty} T_{n_k}(v_i) = L_i$ . So this certainly defines $T$ over all $v_i$ , which is a dense subset of $L^q(E)$ . Is this enough to make $T$ well-defined over all of $L^q(E)$ ? (We're in some danger because the limit in the definition of $T$ might not exist...)

It indeed does! I mean, it's pretty believable. Here are the gory details.

Lemma: Continuity

T is well-defined

T is linear and bounded

Now we need to switch back to functions. By Riesz Representation, there is a unique $u$ such that $T = T_u$ . That is, $T(v) = \int_E uv\,dx$ for all $v$ . At last, we claim that $u_{n_k} \rightharpoonup u$ in $L^p(E)$ . Indeed, for any $v \in L^q(E)$ we have:
$\lim_{k \to \infty} \int_E u_{n_k}v\,dx = \lim_{k \to \infty} T_{n_k}(v) = T(v) = \int_E uv\,dx$ Whew! $\square$

That's one piece of the puzzle down: compactness. The other half, sequential lower-semicontinuity, still needs to be handled. Of course, whether or not we can get sequential lower-semicontinuity will also depend on the functional $F$ , not just which convergence we choose (which, again, will be the weak $L^p$ convergence because it's gotta match what we've been working with above). Naturally, we'll only be able to prove the sequential lower-semicontinuity for some suitably well-behaved functionals.

Researchers in the area, like my advisor Leoni, have proved some different theorems regarding which functionals $F$ will be sequentially lower-semicontinuous with respect to weak convergence. The proofs are completely above my paygrade. In this post, we'll prove a lower-semicontinuity result that arises from convexity. With that...

Intermission: Convexity

You may know a standard definition of convexity from calculus, but we're going to work with a more general notion of convexity.

Definition (Convexity): Let $X$ be a normed space, and let $f:X \to \mathbb{R}$ . We say that $f$ is convex if, for all $x,y \in X$ , we have that
$f((1-t)x+ty) \leq (1-t)f(x) + tf(y)$ for all $t \in [0,1]$ .

"Hey isn't that just Jens-" Yes.

Let's connect this to the notion of convexity that you may be more familiar with.

Theorem: Let $f:\mathbb{R} \to \mathbb{R}$ be twice-differentiable. Then $f$ is convex iff $f''(x) \geq 0$ for all $x$ . For higher dimensions, if $f:\mathbb{R}^N \to \mathbb{R}$ is twice-differentiable, then $f$ is convex iff $\nabla^2 f$ is positive semi-definite.

Proof. See https://math.stackexchange.com/questions/2083629/why-does-positive-semi-definiteness-imply-convexity.

Here's what we actually need though.

Theorem:

Let $\{g_\alpha\}_{\alpha \in \Lambda}$ be a family of affine maps on a normed space $X$ . That is, $g_\alpha(x) = T_\alpha x + c_\alpha$ for some linear $T_\alpha: X \to \mathbb{R}$ and some $c_\alpha \in \mathbb{R}$ . Then $f:X \to \mathbb{R}$ defined by
$f(x) := \sup\left\{g_\alpha(x) : \alpha \in \Lambda\right\}$ is convex.
Conversely (!!!), every convex $f:X \to \mathbb{R}$ may be written as $\sup\left\{g_\alpha(x) : \alpha \in \Lambda\right\}$ for some family $\{g_\alpha\}_{\alpha \in \Lambda}$ of affine maps.

To visualize what this theorem, is saying, take $X = \mathbb{R}$ . Then the motto to imagine is that "you can make any convex graph by drawing a lot of lines".

Proof.

The first point is a boring exercise. The converse is more interesting, and it's what we're actually going to use in the next part.

To proceed, we're going to need to invoke some functional analysis. It's one of those "obvious theorems" that are harder to prove than they look. Essentially: Imagine two convex sets in $\mathbb{R}^2$ . If they don't intersect, then surely you can draw a line separating the two, so that one convex set is on one side of the line and the other convex set is on the other side... right?

The theorem we need is essentially just that but in arbitrarily many dimensions.

Theorem (Geometric Hahn-Banach): Let $Y$ be a normed space, and suppose $A,B \subset Y$ are disjoint, non-empty convex sets, with $A$ open. Then $A$ and $B$ are separated by a closed hyperplane. That is, there exists a continuous linear map $T':Y \to \mathbb{R}$ and a constant $c \in \mathbb{R}$ such that $T'(a) < c \leq T'(b)$ for all $a \in A$ and $b \in B$ .

The "closed hyperplane" that this theorem refers to is basically the set of points $y$ for which $T(y) = c$ .

Why do I want to use this theorem? The key idea is that for $x_0 \in X$ , I want to find an affine function $g_{x_0}$ that passes through $(x_0,f(x_0))$ , such that $g_{x_0}$ just "touches" the graph of $f$ at $x = x_0$ . That is, $g_{x_0}(x_0) = f(x_0)$ , and $g_{x_0}(x) \leq f(x)$ for all $x \in X$ . This indeed sounds like a job for Geometric Hahn-Banach!

Firstly, what is the space $Y$ ? Since I want to think about the "graph" of $f$ , just using the space $X$ won't do. I really want to think about the product space $X \times \mathbb{R}$ , endowed with the norm $\|(x,t)\|_{X \times \mathbb{R}} := \max(\|x\|_X,|t|)$ .

Next, the convex sets $A,B \subseteq Y$ are as follows:

$A$ is the supergraph of $f$ . That is, it is the set of all $(x,t) \in Y$ for which $t > f(x)$ .
$B$ is the ray pointing downward from $(x_0,f(x_0))$ . That is, it is the set $\{(x_0,t) : t \leq f(x_0)\}$ .

We have that $A$ and $B$ are disjoint, and $A$ must be open (why?). Moreover, both $A$ and $B$ are convex (...why?). Thus we may apply the Geometric Hahn-Banach theorem to find a continuous and linear $T':X \times \mathbb{R} \to \mathbb{R}$ and a constant $c$ such that $T'(a) < c \leq T'(b)$ for all $a \in A$ and $b \in B$ .

Now:

$T'(x,t) < c$ for all $(x,t)$ for which $t > f(x)$ . Taking $x = x_0$ , we have $T'(x_0,t) < c$ for all $t > f(x_0)$ . Sending $t \to f(x_0)^+$ , we deduce by continuity of $T'$ that $T'(x_0, f(x_0)) \leq c$ .
$c \leq T'(x_0,t)$ for all $t < f(x_0)$ . Sending $t \to f(x_0)^-$ , we deduce that $c \leq T'(x_0,f(x_0))$ .

Hence $c = T'(x_0,f(x_0))$ .

We are now ready to choose our affine function. The key idea is that we want to take $g_{x_0}(x)$ to be the real number which satisfies $T'(x,g_{x_0}(x)) = c$ (beacuse the hyperplane we want the affine function to represent is given by the set of points where $T' = c$ ). To make this more well-defined, we can manipulate this desired equality into $T'(x,0) + g_{x_0}(x) T'(0,1) = c$ by linearity of $T'$ , and "solving for $g_{x_0}(x)$ " we obtain the definition we want:
$\boxed{g_{x_0}(x) := \frac{c - T'(x,0)}{T'(0,1)}}$ But is this well-defined? We must check that $T'(0,1) \neq 0$ . Indeed, note that $T'(x_0,f(x_0)+0.5) < 0$ and $T'(x_0,f(x_0)-0.5) > 0$ , and subtracting these we get $T'(0,1) < 0$ . Hence we may take this equation to be the definition of $g_{x_0}$ and nobody can complain.

Since $T'(x,0)$ is a continuous linear function $X \to \mathbb{R}$ , we conclude that $g_{x_0}(x)$ is affine. Moreover:
$g_{x_0}(x_0) = \frac{c - T'(x_0,0)}{T'(0,1)} = \frac{T'(x_0,f(x_0)) - T'(x_0,0)}{T'(0,1)}$ $= \frac{T'(0,f(x_0))}{T'(0,1)} = f(x_0)$ So our affine function indeed passes through $(x_0,f(x_0))$ . Hence, for this choice of $g_{x_0}$ to be the one we want, it remains to verify that $g_{x_0}(x) \leq f(x)$ for all $x \in X$ . Manipulating:
$g_{x_0}(x) \leq f(x)$ $\iff \frac{c - T'(x,0)}{T'(0,1)} \leq f(x)$ $\iff c - T'(x,0) \geq f(x)T'(0,1) \qquad (T'(0,1) < 0)$ $\iff c - T'(x,0) \geq T'(0,f(x))$ $\iff c \geq T'(x,f(x))$ And this is indeed true because $T'(x,t) < c$ for all $(x,t)$ for which $t > f(x)$ , and sending $t \to f(x)^+$ we get $T'(x,f(x)) \leq c$ . Brilliant.

To summarize: For every $x_0 \in X$ , we have found an affine $g_{x_0}:X \to \mathbb{R}$ for which $g_{x_0}(x) \leq f(x)$ for all $x \in X$ , and $g_{x_0}(x_0) = f(x_0)$ . I now claim that the family of affine functions for is literally just $\{g_{x_0} : x_0 \in X\}$ . Indeed, the inequality $g_{x_0}(x) \leq f(x)$ holds for all $x \in X$ , and for all $x_0 \in X$ . So we may take the sup to obtain:
$\sup_{x_0 \in X} g_{x_0}(x) \leq f(x)$ But for $x_0 = x$ we obtain equality, hence $\sup_{x_0 \in X} g_{x_0}(x) = f(x)$ for all $x \in X$ . $\square$

Wonderful.

Part 3: Weak Sequential Lower-Semicontinuity

Let's recap: We're looking for a notion of convergence that gets us both a compactness result and a sequential lower-semicontinuity result. $L^p(\Omega)$ convergence is strong enough to get a sequential lower-semicontinuity, but too strong to get a compactness result. Ny switching to weak $L^p(\Omega)$ convergence instead, we managed to get a compactness result. But, is it too weak to get a sequential-lower-semicontinuity result?

For some integral functionals, we might not get weak sequential lower-semicontinuity. However, by imposing a convexity condition, we shall!

Theorem: Let $1 \leq p < \infty$ and let $\Omega \subseteq \mathbb{R}^N$ be open and bounded. Suppose that $f:\mathbb{R} \to \mathbb{R}$ is convex. Then the integral functional $F:L^p(\Omega) \to \overline{\mathbb{R}}$ defined by
$F(u) := \int_\Omega f(u)\,dx$ is sequentially lower-semicontinuous with respect to weak $L^p(\Omega)$ convergence.

Proof.

To properly use our digression into convexity, we first need to prove that...

CLAIM: $F$ is convex.

To see this, take $u_1,u_2 \in L^p(\Omega)$ . Take a $t \in [0,1]$ . Then, by convexity of $f$ , we have that:
$f((1-t)u_1(x) + tu_2(x)) \leq (1-t)f(u_1(x)) + tf(u_2(x)) \qquad \forall x \in \mathbb{R}^N$ Integrating in $x$ :
$\int_\Omega f((1-t)u_1(x) + tu_2(x)) \,dx \leq \int_\Omega (1-t)f(u_1(x)) + tf(u_2(x))\,dx$ $= (1-t)\int_\Omega f(u_1)\,dx + t\int_\Omega f(u_2)\,dx$ But this is just $F((1-t)u_1 + tu_2) \leq (1-t)F(u_1) + tF(u_2)$ . This holds for all $t \in [0,1]$ and all $u_1,u_2 \in L^p(\Omega)$ , hence $F$ is convex.

Alright, now let's prove the theorem. Take a sequence $u_n \in L^p(\Omega)$ with $u_n \rightharpoonup u$ in $L^p(\Omega)$ . What we need to prove is that:
$\liminf_{n \to \infty} F(u_n) \geq F(u)$ By the intermission, and the fact that $F$ is convex, we may write
$F(u) = \sup_{\alpha \in \Lambda} G_\alpha(u)$ for some family of affine functions $G_\alpha:L^p(\Omega) \to \mathbb{R}$ .

What do such affine functions look like? Well, they're just continuous linear maps plus a constant. Hence $G_\alpha = T_\alpha + c_\alpha$ for some continuous linear $T_\alpha : L^p(\Omega) \to \mathbb{R}$ and some $c_\alpha \in \mathbb{R}$ .

But hold on, what does $T_\alpha$ look like? By Riesz-Representation, it must take the form $T_\alpha(u) = \int_\Omega uv_\alpha\,dx$ for some $v_\alpha \in L^q(\Omega)$ , where $q$ is the Holder conjugate of $p$ !

Lit. Let's start by going down easily:
$F(u_n) \geq G_\alpha(u_n) = \int_\Omega u_nv_\alpha\,dx + c_\alpha$ Now let us take the liminf:
$\liminf_{n \to \infty} F(u_n) \geq \liminf_{n \to \infty} \int_\Omega u_nv_\alpha\,dx + c_\alpha$ Hold on, what's going on on the right side? I feel like I should know what $\liminf_{n \to \infty} \int_\Omega u_nv_\alpha\,dx$ is... aha! By the weak convergence $u_n \rightharpoonup u$ , we actually have $\lim_{n \to \infty} \int_\Omega u_nv_\alpha\,dx = \int_\Omega uv_\alpha\,dx$ , because $v_\alpha \in L^q(\Omega)$ ! Therefore:
$\liminf_{n \to \infty} F(u_n) \geq \int_\Omega uv_\alpha\,dx + c_\alpha = G_\alpha(u)$ Finally, as $\alpha \in \Lambda$ was arbitrary, we may take the sup over all indices $\alpha \in \Lambda$ to obtain:
$\liminf_{n \to \infty} F(u_n) \geq \sup_{\alpha \in \Lambda} G_\alpha(u) = F(u)$ Boom. $\square$

Looking back at the functional $F:L^2(\Omega) \to \mathbb{R}$ defined as $F(u) = \int_\Omega u^2\,dx$ from the beginning, we are finally ready to use the direct method to prove that this has a minimum.

Clearly $F(u) \geq 0$ , so it is bounded from below.
Suppose a sequence $u_n \in L^2(\Omega)$ satisfies $\sup_{n \in \mathbb{N}} F(u_n) < \infty$ . Then $u_n$ is bounded in $L^2(\Omega)$ , so by Part 2 there exists a subsequence $u_n \rightharpoonup u \in L^2(\Omega)$ in $L^2(\Omega)$ . So we have a compactness result.
Since $f(z) = z^2$ is convex, we have by the theorem we have just proven that $F$ is sequentially lower-semicontinuous with respect to weak $L^2(\Omega)$ convergence.
Therefore, by the direct method, $F$ has a minimum.

As you can see, we now have the advanced technology necessary to prove that certain functionals of the form $\int_\Omega f(u)\,dx$ for convex $f$ have a minimum! Unfortunately, this isn't really that amazing (why?).

What about this next functional, defined over differentiable functions $u$ ?
$F(u) := \int_0^1 u^2 + (u'-1)^2\,dx$ Hm, the technology we have so far isn't quite advanced enough to deal with derivatives whatsoever...

...we'll talk about how to deal with this in Part 3.

This post has been edited 6 times. Last edited by greenturtle3141, Aug 1, 2022, 6:27 AM

real analysis

analysis

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wait have you considered writing a textbook?

by depsilon0, Aug 1, 2022, 3:30 AM

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At the moment, I don't yet quite consider myself enough of an authority in analysis to confidently write a textbook. However, I'd imagine that it would be fun to do so once I get a PhD or something

I tried to write an olympiad geometry book back in the day, but I never quite finished it.

by greenturtle3141, Aug 1, 2022, 6:25 AM

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Wow you are quite a writer!

by Taco12, Aug 4, 2022, 6:46 PM

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Wait how does your brain know so much
I'm stuggling with speling wile yur dooing

unaleesis

towld yu

by Amkan2022, Feb 21, 2023, 11:00 PM

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Can you give some thought to dropping a guide to STS? Just like how you presented your research (in your paper), what your essays were about, etc. Also cool blog!
by Shreyasharma, Mar 13, 2025, 7:03 PM
this is so good
by purpledonutdragon, Mar 4, 2025, 2:05 PM
orz usamts grader
by Lhaj3, Jan 23, 2025, 7:43 PM
Entertaining blog
by eduD_looC, Dec 31, 2024, 8:57 PM
wow really cool stuff
by kingu, Dec 4, 2024, 1:02 AM
Although I had a decent college essay, this isn't really my specialty so I don't really have anything useful to say that isn't already available online.
by greenturtle3141, Nov 3, 2024, 7:25 PM
Could you also make a blog post about college essay writing :skull:
by Shreyasharma, Nov 2, 2024, 9:04 PM
what gold
by peace09, Oct 15, 2024, 3:39 PM
oh lmao, i was confused because of the title initially. thanks! great read
by OlympusHero, Jul 20, 2024, 5:00 AM
It should be under August 2023
by greenturtle3141, Jul 11, 2024, 11:44 PM
does this blog still have the post about your math journey? for some reason i can't find it
by OlympusHero, Jul 10, 2024, 5:41 PM
imagine not tortoise math

no but seriously really interesting blog
by fruitmonster97, Apr 2, 2024, 12:39 AM
W blog man
by s12d34, Jan 24, 2024, 11:37 PM
very nice blog greenturtle it is very descriptive and fascinating to pay attention to
by StarLex1, Jan 3, 2024, 3:12 PM
orz blog
by ryanbear, Dec 6, 2023, 9:23 PM
Hi! Cool blog!
by mahaler, Nov 14, 2021, 3:08 PM
fascinating...
by pi271828, Nov 14, 2021, 12:00 AM
I believe that indeed, turtles come in all sorts of different colors. Even real ones! Some googling shows that indeed, red turtles seem to exist, though they're not that red.
by greenturtle3141, Nov 13, 2021, 9:21 PM
hello

since you have turtle knowledge, does your species of green turtles have a brethren that possess a color of red? what about blue? what about tawny mauve?
by pi271828, Nov 13, 2021, 9:18 PM
Nice Blog!
by Jwenslawski, Nov 13, 2021, 6:22 PM
Looking especially forward to how you improved on your AMC/AIME/USAMO so quickly
by StopSine, Nov 11, 2021, 2:01 AM
thanks the_turtle
by pog, Nov 10, 2021, 1:09 PM
another turtle blog?
by The_Turtle, Nov 9, 2021, 2:23 AM
Looking forward to future posts!
by Arr0w, Nov 2, 2021, 6:15 PM
just copy and paste every single one of your msm posts and thisll become the best blog ever hahahaha
by peace09, Nov 1, 2021, 9:33 PM
hello!
by tigerzhang, Nov 1, 2021, 9:09 PM
3rd shout! Yay greenturtle blog!
by violin21, Oct 25, 2021, 1:04 AM
hello $\text{[math]}$ $\text{[math]}$
by v4913, Oct 24, 2021, 5:33 PM
first shout
by mathlearner2357, Oct 24, 2021, 5:12 AM

67 shouts

Turtle Math

Calculus of Variations: Part II

by greenturtle3141, Jul 31, 2022, 9:11 PM

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