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conjugate analysis
ironball   6
N Today at 2:04 AM by tohill
Given $f_1,f_2:\mathbb{R}^n\to(-\infty,+\infty]$, prove that $f_1(x)\le f_2(x),\forall x\in\mathbb{R}^n$ if and only if $f_1^*(x)\ge f_2^*(x),\forall x\in\mathbb{R}^n$, where $f^*(x)$ denotes the conjugate function.
6 replies
ironball
Nov 11, 2024
tohill
Today at 2:04 AM
conjugate analysis
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ironball
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Given $f_1,f_2:\mathbb{R}^n\to(-\infty,+\infty]$, prove that $f_1(x)\le f_2(x),\forall x\in\mathbb{R}^n$ if and only if $f_1^*(x)\ge f_2^*(x),\forall x\in\mathbb{R}^n$, where $f^*(x)$ denotes the conjugate function.
This post has been edited 1 time. Last edited by ironball, Nov 11, 2024, 7:04 AM
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ironball
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#2
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bump it up
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tobiSALT
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So we let $f_1,f_2:\mathbb{R}^n\to(-\infty,+\infty]$ be two functions.
The conjugate function of $f: \mathbb{R}^n \to (-\infty, +\infty]$ is defined as
\[f^*(y) = \sup_{x \in \mathbb{R}^n} \{\langle x, y \rangle - f(x)\}\]where $\langle x, y \rangle$ denotes the inner product of $x$ and $y$ in $\mathbb{R}^n$.

We want to prove that $f_1(x) \le f_2(x)$ for all $x \in \mathbb{R}^n$ if and only if $f_1^*(y) \ge f_2^*(y)$ for all $y \in \mathbb{R}^n$.

($\implies$) Suppose $f_1(x) \le f_2(x)$ for all $x \in \mathbb{R}^n$. Then for any $y \in \mathbb{R}^n$, we have
\[f_1^*(y) = \sup_{x \in \mathbb{R}^n} \{\langle x, y \rangle - f_1(x)\}\]and
\[f_2^*(y) = \sup_{x \in \mathbb{R}^n} \{\langle x, y \rangle - f_2(x)\}.\]Since $f_1(x) \le f_2(x)$ for all $x$, we have $-f_1(x) \ge -f_2(x)$ for all $x$. Therefore,
\[\langle x, y \rangle - f_1(x) \ge \langle x, y \rangle - f_2(x)\]for all $x \in \mathbb{R}^n$. Taking the supremum over all $x$, we get
\[\sup_{x \in \mathbb{R}^n} \{\langle x, y \rangle - f_1(x)\} \ge \sup_{x \in \mathbb{R}^n} \{\langle x, y \rangle - f_2(x)\}\]which means $f_1^*(y) \ge f_2^*(y)$ for all $y \in \mathbb{R}^n$.

($\impliedby$) Suppose $f_1^*(y) \ge f_2^*(y)$ for all $y \in \mathbb{R}^n$.
Let $f^{**}(x)$ be the biconjugate of $f(x)$, defined as
\[f^{**}(x) = \sup_{y \in \mathbb{R}^n} \{\langle x, y \rangle - f^*(y)\}.\]It is known that $f^{**}(x) \le f(x)$ for any function $f: \mathbb{R}^n \to (-\infty, +\infty]$.

Since $f_1^*(y) \ge f_2^*(y)$ for all $y \in \mathbb{R}^n$, we have $-f_1^*(y) \le -f_2^*(y)$. Therefore
\[\langle x, y \rangle - f_1^*(y) \le \langle x, y \rangle - f_2^*(y)\]for all $y \in \mathbb{R}^n$. Taking the supremum over all $y$, we get
\[\sup_{y \in \mathbb{R}^n} \{\langle x, y \rangle - f_1^*(y)\} \le \sup_{y \in \mathbb{R}^n} \{\langle x, y \rangle - f_2^*(y)\}\]which means $f_1^{**}(x) \le f_2^{**}(x)$ for all $x \in \mathbb{R}^n$.
If $f_1$ and $f_2$ are closed proper convex functions, then $f_1^{**}(x) = f_1(x)$ and $f_2^{**}(x) = f_2(x)$, so $f_1(x) \le f_2(x)$ for all $x$.

So we get \[f_1(x)\le f_2(x),\forall x\in\mathbb{R}^n \iff f_1^*(x)\ge f_2^*(x),\forall x\in\mathbb{R}^n\]
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ironball
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#4
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tobiSALT wrote:
So we let $f_1,f_2:\mathbb{R}^n\to(-\infty,+\infty]$ be two functions.
The conjugate function of $f: \mathbb{R}^n \to (-\infty, +\infty]$ is defined as
\[f^*(y) = \sup_{x \in \mathbb{R}^n} \{\langle x, y \rangle - f(x)\}\]where $\langle x, y \rangle$ denotes the inner product of $x$ and $y$ in $\mathbb{R}^n$.

We want to prove that $f_1(x) \le f_2(x)$ for all $x \in \mathbb{R}^n$ if and only if $f_1^*(y) \ge f_2^*(y)$ for all $y \in \mathbb{R}^n$.

($\implies$) Suppose $f_1(x) \le f_2(x)$ for all $x \in \mathbb{R}^n$. Then for any $y \in \mathbb{R}^n$, we have
\[f_1^*(y) = \sup_{x \in \mathbb{R}^n} \{\langle x, y \rangle - f_1(x)\}\]and
\[f_2^*(y) = \sup_{x \in \mathbb{R}^n} \{\langle x, y \rangle - f_2(x)\}.\]Since $f_1(x) \le f_2(x)$ for all $x$, we have $-f_1(x) \ge -f_2(x)$ for all $x$. Therefore,
\[\langle x, y \rangle - f_1(x) \ge \langle x, y \rangle - f_2(x)\]for all $x \in \mathbb{R}^n$. Taking the supremum over all $x$, we get
\[\sup_{x \in \mathbb{R}^n} \{\langle x, y \rangle - f_1(x)\} \ge \sup_{x \in \mathbb{R}^n} \{\langle x, y \rangle - f_2(x)\}\]which means $f_1^*(y) \ge f_2^*(y)$ for all $y \in \mathbb{R}^n$.

($\impliedby$) Suppose $f_1^*(y) \ge f_2^*(y)$ for all $y \in \mathbb{R}^n$.
Let $f^{**}(x)$ be the biconjugate of $f(x)$, defined as
\[f^{**}(x) = \sup_{y \in \mathbb{R}^n} \{\langle x, y \rangle - f^*(y)\}.\]It is known that $f^{**}(x) \le f(x)$ for any function $f: \mathbb{R}^n \to (-\infty, +\infty]$.

Since $f_1^*(y) \ge f_2^*(y)$ for all $y \in \mathbb{R}^n$, we have $-f_1^*(y) \le -f_2^*(y)$. Therefore
\[\langle x, y \rangle - f_1^*(y) \le \langle x, y \rangle - f_2^*(y)\]for all $y \in \mathbb{R}^n$. Taking the supremum over all $y$, we get
\[\sup_{y \in \mathbb{R}^n} \{\langle x, y \rangle - f_1^*(y)\} \le \sup_{y \in \mathbb{R}^n} \{\langle x, y \rangle - f_2^*(y)\}\]which means $f_1^{**}(x) \le f_2^{**}(x)$ for all $x \in \mathbb{R}^n$.
If $f_1$ and $f_2$ are closed proper convex functions, then $f_1^{**}(x) = f_1(x)$ and $f_2^{**}(x) = f_2(x)$, so $f_1(x) \le f_2(x)$ for all $x$.

So we get \[f_1(x)\le f_2(x),\forall x\in\mathbb{R}^n \iff f_1^*(x)\ge f_2^*(x),\forall x\in\mathbb{R}^n\]

Thank you for your great answer! But what if f_1,f_2 are not proper closed convex functions?
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Gryphos
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#5
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ironball wrote:
Given $f_1,f_2:\mathbb{R}^n\to(-\infty,+\infty]$, prove that $f_1(x)\le f_2(x),\forall x\in\mathbb{R}^n$ if and only if $f_1^*(x)\ge f_2^*(x),\forall x\in\mathbb{R}^n$, where $f^*(x)$ denotes the conjugate function.
This is false without further assumptions. Indeed, take e.g. $f_2(x) = x^2$ and
$$f_1(x) = \begin{cases} x^2 & \colon x \ne 0
\\
1 & \colon x = 0.
\end{cases}$$It is not hard to see that $f_1^*(x) = f_2^*(x) = \textstyle\frac{x^2}{4}$, but $f_1(0) > f_2(0)$.
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tobiSALT
28 posts
#6
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ironball wrote:
tobiSALT wrote:
So we let $f_1,f_2:\mathbb{R}^n\to(-\infty,+\infty]$ be two functions.
The conjugate function of $f: \mathbb{R}^n \to (-\infty, +\infty]$ is defined as
\[f^*(y) = \sup_{x \in \mathbb{R}^n} \{\langle x, y \rangle - f(x)\}\]where $\langle x, y \rangle$ denotes the inner product of $x$ and $y$ in $\mathbb{R}^n$.

We want to prove that $f_1(x) \le f_2(x)$ for all $x \in \mathbb{R}^n$ if and only if $f_1^*(y) \ge f_2^*(y)$ for all $y \in \mathbb{R}^n$.

($\implies$) Suppose $f_1(x) \le f_2(x)$ for all $x \in \mathbb{R}^n$. Then for any $y \in \mathbb{R}^n$, we have
\[f_1^*(y) = \sup_{x \in \mathbb{R}^n} \{\langle x, y \rangle - f_1(x)\}\]and
\[f_2^*(y) = \sup_{x \in \mathbb{R}^n} \{\langle x, y \rangle - f_2(x)\}.\]Since $f_1(x) \le f_2(x)$ for all $x$, we have $-f_1(x) \ge -f_2(x)$ for all $x$. Therefore,
\[\langle x, y \rangle - f_1(x) \ge \langle x, y \rangle - f_2(x)\]for all $x \in \mathbb{R}^n$. Taking the supremum over all $x$, we get
\[\sup_{x \in \mathbb{R}^n} \{\langle x, y \rangle - f_1(x)\} \ge \sup_{x \in \mathbb{R}^n} \{\langle x, y \rangle - f_2(x)\}\]which means $f_1^*(y) \ge f_2^*(y)$ for all $y \in \mathbb{R}^n$.

($\impliedby$) Suppose $f_1^*(y) \ge f_2^*(y)$ for all $y \in \mathbb{R}^n$.
Let $f^{**}(x)$ be the biconjugate of $f(x)$, defined as
\[f^{**}(x) = \sup_{y \in \mathbb{R}^n} \{\langle x, y \rangle - f^*(y)\}.\]It is known that $f^{**}(x) \le f(x)$ for any function $f: \mathbb{R}^n \to (-\infty, +\infty]$.

Since $f_1^*(y) \ge f_2^*(y)$ for all $y \in \mathbb{R}^n$, we have $-f_1^*(y) \le -f_2^*(y)$. Therefore
\[\langle x, y \rangle - f_1^*(y) \le \langle x, y \rangle - f_2^*(y)\]for all $y \in \mathbb{R}^n$. Taking the supremum over all $y$, we get
\[\sup_{y \in \mathbb{R}^n} \{\langle x, y \rangle - f_1^*(y)\} \le \sup_{y \in \mathbb{R}^n} \{\langle x, y \rangle - f_2^*(y)\}\]which means $f_1^{**}(x) \le f_2^{**}(x)$ for all $x \in \mathbb{R}^n$.
If $f_1$ and $f_2$ are closed proper convex functions, then $f_1^{**}(x) = f_1(x)$ and $f_2^{**}(x) = f_2(x)$, so $f_1(x) \le f_2(x)$ for all $x$.

So we get \[f_1(x)\le f_2(x),\forall x\in\mathbb{R}^n \iff f_1^*(x)\ge f_2^*(x),\forall x\in\mathbb{R}^n\]

Thank you for your great answer! But what if f_1,f_2 are not proper closed convex functions?

Then it doesn´t hold as the guy above said which is easy to see, I thought about it in the hard way lol.
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tohill
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cool problem
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