Can we approximate a convex function by twice differentiable convex function?

Question

Suppose $f(x)$ is convex over some compact set $S \subset \mathbb{R}^n$. Can we construct a sequence $f_n$ of twice differentiable convex function such that:

1. $f_n(x) \rightarrow f(x)$ for all $x \in S$.
2. $f_n'(x) \rightarrow f'(x)$ for all $x \in S$ where $f'(x)$ exists.
3. $f_n''(x) \rightarrow f''(x)$ for all$x \in S$ where $f''(x)$ exists.

Additionally, it would be helpful if the convergence is uniform.

Motivation: I am reading a textbook on convex optimization, and there is a variety of inequalities on convex functions proved under various assumptions (simply convex, convex+differentiable, twice differentiable, etc). I'm wondering if one can just assume everything is twice differentiable, derive the needed results, and argue they hold for non-twice-differentiable functions by a limiting argument.

Answer 1

Uniform convergence will not be possible, I think. Roughly speaking, suppose that $f'(x)$ has a discontinuity, but $f''(x)$ is bounded at all points except this discontinuity. The functions $f_n'(x)$ will have to be "nearly discontinuous", converging to one value above the discontinuity and another below it. In other words, the functions $f'_n(x)$ will have to change rapidly in a small neighborhood of the discontinuity. But this will lead to $f''_n(x)$ being quite large in that same neighborhood, and so $f''_n(x)$ will "have trouble" converging uniformly to a finite function $f''(x)$.

Here's my attempt at a proof. It's probably not as rigorous as it should be, and I welcome any critiques.

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Consider the function $f(x) = |x|$ from $[-1,1] \to [0,1]$. Suppose that a family of functions $f_n(x)$ exists such that $f'_n(x)$ and $f''_n(x)$ converge absolutely to $f'(x)$ and $f''(x)$ where they are defined (i.e., on $[-1,0) \cup (0,1]$.

Fix two positive numbers $0<a<1$ and $0 < \epsilon < 1/(a+1)$. We have that $f''(x) = 0$, where it exists. Uniform convergence then implies that there exists an $N$ such that for $n > N$, $|f_n''(x)| < \epsilon$ for $x \neq 0$. This then implies that
$$ |f'_n(a) - f'_n(-a)| = \left| \int_{-a}^a f''_n(x) dx \right| \leq \int_{-a}^a |f''_n(x)| dx < 2 a \epsilon $$ for all $n > N$.

But if $f'_n(x)$ is to converge uniformly to $f'(x)$ on $[-1,0) \cup (0,1]$, then for any $\epsilon$ there must also exist an $M$ such that for all $n > M$, we have $|f'_n(x) - f'(x)| < \epsilon$. For any $a> 0$ we have $f'(a) = 1$ and $f'(-a)= -1$. This means, in particular, that for $n > M$ we must have $|f'_n(a) - 1| < \epsilon$ and $|f_n(-a) + 1| < \epsilon$; together, these imply that $$ |f'_n(a) - f'_n(-a)| > 2 - 2 \epsilon. $$

Thus, for any $n > \max(N,M)$, we must have $$ 2 a \epsilon > |f'_n(a) - f'_n(-a)| > 2 - 2 \epsilon $$ which implies that $\epsilon > 1/(a+1)$. But $\epsilon$ was chosen such that $\epsilon < 1/(a+1)$, and so we have a contradiction.

Answer 2

Surely we can! Let $$f_n(x)=f(x)-\dfrac{x+1}{n}$$for example.