Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I want to show that if $X$ is a reflexive Banach space with norm of class $\mathcal{C}^1$ and $f\colon X\to\mathbb{R}\cup \{+\infty\}$ is convex and lower semicontinuous, then $f_{\lambda}$ is differentiable of class $\mathcal{C}^1$.

(where $f_{\lambda}:X\to\mathbb{R}\cup \{+\infty\}$ is the Moreau-Yosida approximation: $$f_\lambda(x)=\inf_{y\in X} \left\{ f(y)+\frac{1}{2\lambda}|x-y|^2\right\})$$

Maybe, this result could be useful: If $g\colon X\to\mathbb{R}$ is convex and differentiable in every point then $g\in\mathcal{C}^1(X)$.

Many thanks in advance.

share|improve this question
    
Do you know if this works when $X$ is Hilbert space? And if so, would the argument generalize to other uniformly convex spaces? –  user16299 Jan 15 '12 at 3:09
    
What does $\mathcal{C}^1$ mean in this context? –  Simen K. Feb 7 '13 at 10:23
add comment

1 Answer 1

First, a remark: a convex lower semicontinuous function is weakly lower semicontinuous. Indeed, lower semicontinuity is equivalent to the epigraph being closed in the appropriate topology. The epigraph of a convex function is convex. A closed convex set is weakly closed.

Fix $\lambda>0$. For any $y$, the function $g_{y}(x)= f(y)+\frac{1}{2\lambda}|x-y|^2$ is convex. Taking infimum on both sides of $g_y((1-t)a+tb)\le (1-t)g_y(a)+tg_y(b)$, we find that $f_\lambda$ is convex.

Now fix $x$ and pick a sequence $(y_n)$ such that $g_{y_n}(x)\to f_\lambda(x)$. It is not hard to see that $g_y(x)\to +\infty$ when $\|y\|\to \infty$. Therefore, the sequence $(y_n)$ is bounded. Since we are in a reflexive space, there is a weakly convergent subsequence $y_{n_k}\to z$. Since $f$ is weakly lower semicontinuous (and so is the squared norm), it follows that $g_z(x) \le f_\lambda(x)$. Thus, the infimum in the definition of $f_\lambda$ is attained: $g_z(x) = f_\lambda(x)$.

For any other point $x' $ we have $f_\lambda(x')\le g_z(x') $. Hence $f_\lambda(x')-f_\lambda(x) \le g_z(x')-g_z(x) $. Since the squared norm is differentiable, so is $g_z$. As $x'\to x$, we find that $f_\lambda(x')-f_\lambda(x) \le \varphi(x'-x)+o(1)$ where $\varphi\in X^*$ is the derivative of $g_z$ at $x$. On the other hand, convexity implies $f_\lambda(x')-f_\lambda(x)\ge \psi(x'-x)$ for some $\psi\in X^*$. It follows that $\varphi=\psi$, and this functional is the Fréchet derivative of $f_\lambda$ at $x$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.