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.

  • $\begingroup$ Do you know if this works when $X$ is Hilbert space? And if so, would the argument generalize to other uniformly convex spaces? $\endgroup$
    – user16299
    Jan 15, 2012 at 3:09
  • $\begingroup$ What does $\mathcal{C}^1$ mean in this context? $\endgroup$
    – Jas Ter
    Feb 7, 2013 at 10:23

2 Answers 2


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 $x$.


Observe that the subdifferential of the function $y\to \frac{\|x-y\|^2}{2\lambda}+f(y)$ is the operator $$y\to F(y-x)+\partial f(y),$$ where $F:X\to X^*$ is a duality mapping ($Fx=\{f^*\in X^*\,|\,\langle f,x\rangle=\|x\|^2=\|f\|^2\}$). Now, recall that a point $y$ is a minimizer of a convex function $g$ if and only if $0\in \partial g(y)$. Since $F$ is a duality mapping, $A=\partial f$ is maximal monotone and $X$ is reflexive, invoking Rockafellar (or Minty - don't remember) theorem, we have that the equation $$F(y-x)+Ay\ni0$$ has a unique solution, which gives that the infimum is attained. Contrary to user53153 answer, the argument which realizes the infimum is not only a weak limit of a minimizing sequence, but also a solution to a certain equation. This has a direct impact on Gâteaux differentiability of $f_\lambda$ if we don't assume that the norm is $\mathcal{C}^1$.


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