Set-up: Let $\{f_n \}_{n=1}^\infty$ be a sequence of smooth convex functions on $(0,1) \subset \mathbb R$ that converge uniformly to the continuous (not necessarily differentiable) convex function $f$.

It is well known that $f$ is differentiable almost everywhere. It is also known that $\{f'_n \}$ converges pointwise to $f'$ wherever $f'$ exists.

Further, by Alexandrov's theorem, $f$ is also twice differentiable almost everywhere.

My question: Is there any type of convergence of $2$nd derivatives?

Thoughts: Clearly, there is not uniform convergence since at any point where $f'$ does not exists, $f''$ will be infinite (e.g., $f(x) =|x|$). But maybe pointwise a.e. or convergence in measure?


1 Answer 1


Bad news

The second derivatives need not converge a.e. or even in measure. Indeed, let $g_n(x)$ be the $n$th binary digit of $x$. Then the antiderivatives $G_n(x)=\int_0^x g_n(t)\,dt$ uniformly converge to $x/2$, and second antiderivatives $f_n(x)=\int_0^x G_n(t)\,dt$ uniformly converge to $f(x)=x^2/4$. However, $|g_n-f''|=1/2$ a.e.

(In this example, $f_n$ is not twice differentiable, but one can fix this by smoothing $g_n$ a little; this doesn't really matter for integral estimates.)

Good news

Distributional convergence is inherited by derivatives of all orders. This means that for any test function $\phi\in C_c^\infty((0,1))$ we have $$ \int_0^1 f_n'' \phi\,dx \to \int_0^1 f''\phi\,dx \tag{1}$$ This is not particularly surprising: it's just a restatement of the obvious $$ \int_0^1 f_n \phi''\,dx \to \int_0^1 f \phi''\,dx $$
via integration by parts.

But we can do better: (1) holds for all continuous functions $\phi$ with compact support in $(0,1)$. In other words: interpreting $f_n''$ as a measure (which it is), we have weak convergence of measures on compact subsets of $(0,1)$. Here is why:

  1. The measures are bounded on each interval $[a,b]$ with $0<a<b<1$, since $f_n'(a)$ and $f_n'(b)$ are uniformly bounded (controlled by convexity & uniform convergence).
  2. By Banach-Alaoglu, every subsequence $f_{n_k}''$ has a weakly convergent subsequence $f_{n_{k_j}}''$.
  3. Since weak convergence of measures implies distributional convergence, the limit of $f_{n_{k_j}}''$ is $f''$.
  4. General topology fact: if every subsequence has a subsubsequence converging to the same limit, then the whole sequence converges there.

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