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I was wondering if the fenchel conjugate of the $\frac{1}{2}||u||^2$, is the $\frac{1}{2}||u||_*^2$, where $||.||_*$ is the dual norm of $||.||$. This seems to be true for the $\ell_2$ norm. However, I do not seem to be able to prove it in general.

Does anyone know if this is even true in general?

Devil
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2 Answers2

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For convenience, I'll transcribe (word for word) the proof given in Example 3.27 (pp. 93-94) of Boyd and Vandenberghe (which is free online). Note that in the passage below, the domain of $f$ is $\mathbb R^n$.

Now consider the function $f(x) = (1/2) \| x \|^2$, where $\| \cdot \|$ is a norm, with dual norm $\| \cdot \|_*$. We will show that its conjugate is $f^*(y) = (1/2) \|y\|_*^2$. From $y^T x \leq \| y \|_* \|x\|$, we conclude $$ y^T x - (1/2) \| x \|^2 \leq \| y \|_* \| x \| - (1/2) \|x\|^2 $$ for all $x$. The righthand side is a quadratic function of $\|x\|$, which has maximum value $(1/2)\|y\|_*^2$. Therefore for all $x$, we have $$ y^T x - (1/2) \|x\|^2 \leq (1/2) \|y\|_*^2 $$ which shows that $f^*(y) \leq (1/2) \|y\|_*^2$.

To show the other inequality, let $x$ be any vector with $y^T x = \|y\|_* \|x\|$, scaled so that $\|x\| = \|y\|_*$. Then we have, for this $x$, $$ y^T x - (1/2) \|x\|^2 = (1/2) \|y\|_*^2, $$ which shows that $f^*(y) \geq (1/2) \|y\|_*^2$.

littleO
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  • This proof seems to assume that there exists, for each $y\in X^$, an element $x\in X$ such that $\langle y, x \rangle = ||y||_||x||$. By James's theorem this is only true if $X$ is reflexive, right? Wouldn't it be better to take an approximating sequence $x_n$ to make the proof rigorous? – BigbearZzz Feb 04 '19 at 15:46
  • @BigbearZzz In this book $X$ is $\mathbb R^n$, so the book doesn't need to worry about that issue. – littleO Feb 04 '19 at 16:14
  • Ah, I see. However the OP seemed to ask about general Banach space since he mentioned $l^2$. I think it might be better if you perhaps make a remark about this in the answer? – BigbearZzz Feb 04 '19 at 16:17
  • @BigbearZzz I just updated the answer. But, even in a finite dimensional setting, optimization is very interested in other norms such as the $\ell_1$ and $\ell_\infty$ norms. – littleO Feb 04 '19 at 16:24
  • I think you are right. From my background, I automatically think about the infinite dimensional Hilbert space when I saw the symbol $\ell_2$. It didn't occur to me at all that the OP may talk about the $\ell_2$ norm on $\Bbb R^n$. – BigbearZzz Feb 04 '19 at 16:27
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Yes, this is true. Please see page 93-94 in the book "convex optimization" https://web.stanford.edu/~boyd/cvxbook/bv_cvxbook.pdf

Yunwen Lei
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