Continuous function and maximum value

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Continuous function and maximum value

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Koukai

#1 h Mar 26, 2015, 5:48 PM • 1 Y

Y by Adventure10

Let $E$ be the set of all continuous functions $u:[0,1]\longmapsto\mathbb R$ that satisfy $|u(x)-u(y)|\le|x-y|$ for $0\le x,y\le1$ and $u(0)=0.$ Let $f:E\longmapsto\mathbb R$ be defined by $f(u)=\int_0^1(u^2(x)-u(x))\,dx.$ Show that $f$ attains its maximum value at some element of $E.$

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WWW

#2 h Mar 26, 2015, 9:40 PM • 2 Y

Y by Adventure10, Mango247

We have $|u(x)|\le x,x\in [0,1],$ for all $u \in E.$ So for all $u\in E,$ $\int_0^1(u^2-u) \le \int_0^1(u^2+|u|) \le \int_0^1(x^2+x)\,dx = \frac{5}{6}.$ This value is attained when $u(x)=-x,$ which belongs to $E,$ so we're done.

This post has been edited 2 times. Last edited by WWW, Mar 26, 2015, 9:43 PM

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Koukai

#3 h Mar 26, 2015, 10:38 PM • 2 Y

Y by Adventure10, Mango247

Thank you.
When $u(x)=-x,$ it means that $\int_0^1(u^2-u)\,dx=\int_0^1(u^2+|u|)\,dx?$

This post has been edited 1 time. Last edited by Koukai, Mar 26, 2015, 10:39 PM

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WWW

#4 h Mar 26, 2015, 11:39 PM • 2 Y

Y by Adventure10, Mango247

You don't need me to answer that.

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jmerry

#5 h Mar 26, 2015, 11:53 PM • 3 Y

Y by Koukai, Adventure10, Mango247

All right, now the ridiculous way:

First, note that given any compact topological space $K$ , a continuous function from $K$ to $\mathbb{R}$ attains its maximum and minimum - this is a version of the Extreme Value Theorem, which follows from the fact that the continuous image of a compact set is compact. We wish to show that under some reasonable topology, $E$ is compact and $f$ is continuous.

To define the topology on $E$ , let's try a natural choice - the uniform norm $\|u\|_{\infty}=\sup_x |u(x)|$ on bounded functions. The distance between two functions $u$ and $v$ is then $\|u-v\|_{\infty}$ , and it is fairly easy to verify the axioms of a metric space. We do need to note that $\|u(x)\|\le 1$ for all $u\in E$ and all $x$ , so everything in our space has a norm. Since our $u$ are continuous, the supremum in the definition is actually a maximum.

Now, what does that $|u(x)-u(y)|\le |x-y|$ get us? Since that applies everywhere in the interval, that uniform (Lipschitz) continuity. Since it applies to all $u$ in $E$ , that's also equicontinuity. $E$ is a uniformly bounded and equicontinuous family of functions on a compact interval... so we can apply the Arzela-Ascoli Theorem. Any sequence in $E$ has a uniformly convergent subsequence; if we can show that the limit of such a subsequence is always in $E$ , we have that $E$ is compact.

So, the final lemma: if $\|u_n-u\|_{\infty}\to 0$ with each $u_n\in E$ , $u$ is also in $E$ .
For any $\epsilon>0$ , there is some $N$ such that $|u_n-u|_{\infty}<\epsilon$ whenever $n>N$ . Choose $x$ and $y$ , as well as some particular $n>N$ ; we then have $|u(x)-u(y)|\le |u(x)-u_n(x)|+|u_n(x)-u_n(y)|+|u_n(y)-u(y)|< 2\epsilon+|x-y|$ . Now, since $\epsilon$ was arbitrary, we let it go to zero; if $|u(x)-u(y)|<2\epsilon+|x-y|$ for all $\epsilon>0$ , $|u(x)-u(y)|\le |x-y|$ . That's the definition of $E$ , and we're done.

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Koukai

#6 h Mar 27, 2015, 12:36 AM • 1 Y

Y by Adventure10

WWW wrote:

You don't need me to answer that.

Okay, it was obvious.

Thanks jmerry for the details.

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jmerry

#7 h Mar 27, 2015, 1:33 AM • 3 Y

Y by Koukai, Adventure10, Mango247

The details? What I wrote was a completely different argument, which doesn't say anything about where the maximum is. I did call it the ridiculous way.

Actually, there's one detail I forgot: showing that $f$ is continuous. We have $(u^2(x)-u(x))-(v^2(x)-v(x))=(u(x)-v(x))(u(x)+v(x)+1)$ , so $\left|(u^2(x)-u(x))-(v^2(x)-v(x))\right|\le 3(u(x)-v(x))\le 3\|u-v\|_{\infty}$ . Integrate that and $|f(u)-f(v)|\le 3\|u-v\|_{\infty}$ . Good enough.

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