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I'm trying to prove this well-known approximation of proper l.s.c. convex function. Could you have a check on my attempt?

Let $(X, | \cdot|)$ be a normed space and $f:X \to \mathbb R \cup \{+\infty\}$ convex proper lower semi-continuous. Then there is an increasing sequence of Lipchitz-continuous convex functions $(f_n)$ with $f_n:X \to \mathbb R$ such that $f_n \nearrow f$ pointwise.

My attempt: We need the following lemma.

Lemma: Any convex proper lower semi-continuous function is bounded from below by a continuous affine function.

We define $f_n:X \to \mathbb R \cup \{\pm\infty\}$ by $$ f_n (x) := \inf_{y\in X} \{ f(y) + n|x-y| \} \quad \forall x\in X. $$

Clearly, $f_n$ is convex and $f_n \le f_{n+1} \le f$. We have $$ \begin{align} f_n (x) &= \inf_{y\in X}\{f(y)+n |x-y|\} \\ &\le \inf_{y\in X}\{f(y)+n |x-z| + n |y-z|\} \\ &= f_n(z)+n |x-z|. \end{align} $$ By symmetry, we get $|f_n(x)-f_n(z)|\ \le n |x-z|$. Thus $f_n$ is $n$-Lipschitz continuous.

Because $f$ is proper, there is $a \in X$ such that $f(a) < +\infty$. Then $f_n (x) \le f(a) + n|a-x| <+\infty$ for all $x\in X$. By our Lemma, there is a linear continuous functional $g$ and $\alpha \in \mathbb R$ such that $g +\alpha \le f$. Then $$ \begin{align} f(y) + n|x-y| &\ge g(y)+n|x-y| +\alpha \\ &\ge -\|g\||y| + n|x-y|+\alpha \\ &\ge (n-\|g\|) |y| -n |x| +\alpha. \end{align} $$

If $n \ge \|g\|$, then $f_n (x) \ge-n|x| +\alpha > -\infty$ for all $x\in X$. Clearly, $(f_n)$ is increasing. WLOG, we re-index $(f_n)$ such that $f_n(x) \in \mathbb R$ for all $n \in \mathbb N$ and $x \in X$.

Let's prove that $f_n (x) \to f(x)$.

  • Fix $x \in \operatorname{dom} f$. We pick $x_n \in X$ such that $$ f(x_n)+n|x-x_n|<f_n(x)+\frac{1}{n}. $$

First, $f_n(x) \le f(x)$. Second, $f(x_n) \ge g(x_n)+\alpha \ge -\|g\||x_n|+\alpha$, so $$ n|x-x_n| < f(x)+1/n+\|g\||x_n|-\alpha. $$ This implies $|x_n-x| \to 0$. Hence $$ \lim_n f_n(x) \ge \lim_n f(x_n) \ge f(x). $$

  • Fix $x \not\in \operatorname{dom} f$. Then $f(x) = +\infty$. We claim that $\lim_n f_n(x) = +\infty$. Assume the contrary that $(f_n(x))$ is bounded by $C <+\infty$. We pick $x_n \in X$ such that $$ f(x_n)+n|x-x_n|<f_n(x)+\frac{1}{n}. $$

Then $$ f(x_n)+n|x-x_n| \le C+1/n. $$

Just as above, we get $|x-x_n| \to 0$. Then $f(x) \le \liminf_n f(x_n) \le C$, which is a contradiction. This completes the proof.

Analyst
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  • The problem with $f_n$ as you defined it is that it may take the value $-\infty$ in which case, you would have situations in where $|f_n(x)-f_n(y)|$ is wither not defined ($-\infty-(-\infty)$) to takes the value $\infty$ ($-\infty-real$). I made a modification to the proof to fix that problem. The argument then follows close to what you did. – Mittens Feb 04 '25 at 19:19

1 Answers1

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There is a faulty argument in that $|f_n(x)-f_n(y)|\leq n\|x-y\|$ since it may happen that $f_n(y)=-\infty$ or $-\infty=f_n(x)$ or both.

The proof can be fixed as follows. Choose any continuous affine function $g$ such that $g(y)<f(y)$ for all $y\in X$ (see here). The function $\phi(y)=f(y)-g(y)$ is also l.s.c and convex. The advantage of this is that $\phi$ is bounded below: $\phi(y)>0$ for all $y\in X$.

The arguments in proof are now as in the OP's. For $t>0$ define $$f_t(y)=\inf_{z\in X}\{\phi(z)+t\|y-z\|\}$$ It is clear that $f_t$ is convex and that $$0\leq f_s(y)\leq f_t(y)\leq \phi(y)\qquad 0<s<t,\quad y\in X.\tag{0}\label{zero}$$ Furthermore, as $\operatorname{dom}(f)\neq\emptyset$, $f_t(y)\in[0,\infty)$. From the triangle inequality, we also have that $$f_t(y)\leq f_t(x)+t\|x-y\|\qquad\forall (x,y)\in X\times X$$ Therefore, $$|f_t(y)-f_t(x)|\leq t\|x-y\|\qquad\forall (x,y)\in X\times X\tag{1}\label{one}$$ In other words, the functions $f_t$ are Lipschitz continuous. From \eqref{one}, $f(y):=\lim_{t\rightarrow\infty}f_y(y)\leq \phi(y)$. we now prove that $f=\phi$. Fix $x\in X$ and let $\varepsilon>0$. For each $n\in\mathbb{N}$ there exists $z_n\in X$ such that $$ \min\left\{\phi(x)+\varepsilon,f(x)+\varepsilon\right\}\geq f_n(x)+\varepsilon>\phi(z_n)+n\|x-z_n\|> \max\left\{\phi(z_n),n\|z-z_n\|\right\}\tag{3}\label{three}$$ If $x\in\operatorname{dom}(f)$, then $\phi(x)\in(0,\infty)$ and so, $\|z_n-x\|\xrightarrow{n\rightarrow\infty}0$. Since $\phi$ is l.s.c, there is $N$ such that $$\phi(x)-\varepsilon < \phi(z_n)\qquad \forall n\geq N\tag{4}\label{four}$$ Combining \eqref{three} and \eqref{four} $$f_n(x)>\phi(x)-2\varepsilon$$ It follows that $$\phi(x)\geq f(x)=\lim_nf_n(x)\geq\phi(x)-2\varepsilon$$ for all $\varepsilon>0$; thus, $f(x)=\phi(x)$.

It remains to consider the case where $\phi(x)=\infty$. If $f(x)=\infty$ there is nothing to do. If $f(x)<\infty$, then $\|z_n-x\|\xrightarrow{n\rightarrow\infty}0$. Since $\phi$ is l.s.c., for any $a>0$ there is $N$ such that $$a<\phi(z_n)\qquad \forall n\geq N$$ By \eqref{three} $$a<\phi(z_n)<f_n(x)+\varepsilon\leq f(x)+\varepsilon $$ This yields a contradiction for we may take $a>f(x)+\varepsilon$. Therefore, $f(x)=\infty$.

The proof above shows that there exists a monotone increasing sequence of Liptchitz functions $f_n$ that converge pointwise to $f-g$. The sequence $\psi_n=f_n+g$ is also Lipschitz continuous and converges to $f$.

Mittens
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