There is a well known theorem on the relationship between uniform convergence of univariate functions and differentiation. Quoting Theorem 7.17 from Rudin 1976, Principles of Mathematical Analysis:

Theorem: Suppose $\{f_n\}$ is a sequence of functions, differentiable on $[a, b]$ and such that $\{f_n(x_0)\}$ converges for some point $x_0$ on $[a,b]$. If $\{f_n'\}$ converges uniformly on $[a,b]$, then $\{f_n\}$ converges uniformly on $[a,b]$ to a function $f$, and $f'(x) = \lim_{n\rightarrow\infty} f_n'(x)$.

My question is, does a generalisation of this theorem exist for sequences of functions $f_n: \mathbb{R}^n \rightarrow \mathbb{R}?$ For example, can $f'_n$ be replaced by $\nabla f_n$ and $[a,b]$ be replaced by a compact set?


1 Answer 1


After some thought, I have this corollary to the theorem quoted in the question.

Corollary: Let $f_n : \mathcal{X} \rightarrow \mathbb{R}$ be a family of functions on a subset $\mathcal{X} \subseteq \mathbb{R}^n$ and suppose $\{f_n(x)\}$ converges pointwise to a function $f$ on a convex set $\mathcal{C} \subseteq \mathcal{X}$. Further, suppose that $\nabla f_n$ converges uniformly on $\mathcal{C}$. Then $\{f_n\}$ converges uniformly on $\mathcal{C}$ and $\nabla f = \lim_{n\rightarrow \infty} \nabla f_n$.

Proof: Fix two points $a, b \in \mathcal{C}$. By convexity of $\mathcal{C}$, all points $g(t)$ on the straight line connecting $a$ and $b$ are in $\mathcal{C}$: $$ g(t) = (1-t)a + tb \in \mathcal{C}\; \forall\; t\in[0,1]. $$ Then we have $f_n \circ g$ pointwise convergent on $t\in[1,0]$. Furthermore, $(\nabla f_n \cdot (b-a))\circ g$ is uniformly convergent on $t\in[0,1]$. Indeed, fixing $\epsilon > 0$, by the Cauchy-Schwarz inequality, we have $$ |(\nabla f_n (x) - \nabla f_m (x))\cdot(b-a)| \leq ||b-a||\,||\nabla f_n(x) - \nabla f_m(x)||, $$ and by uniform convergence of $\nabla f_n$, there is some $N(\epsilon)$ such that, whenever $n > N$ and $m > N$, the right hand side is less than $||b-a||\epsilon$. It is quick to verify that $(\nabla f_n \cdot (b-a))\circ g = \partial f_n \circ g/\partial t$. It follows from the theorem stated in the question that $f_n \circ g(t)$ converges uniformly to $f\circ g(t)$ on $[0, 1]$ and $$ \frac{\partial}{\partial t}f \circ g(t) = \lim_{n\rightarrow \infty} \frac{\partial}{\partial t} f_n \circ g(t). $$ As the choice of $a$ and $b$ is arbitrary, this shows that $\nabla f_n\cdot l \rightarrow \nabla f \cdot l$ in all directions $l$. Choosing $l$ to be the basis vectors, this shows $\nabla f_n \rightarrow \nabla f$.

  • $\begingroup$ Thanks for the answer. Just one comment: in your corollary, I think you may want the domain of $f_n$ to be some subset of $\mathbb{R}^d$; metric space is not enough to talk about derivatives. $\endgroup$
    – Jiaqi Li
    Commented Feb 28, 2018 at 17:55
  • $\begingroup$ Thanks Jiaqi Li, good point. I'll update the answer. $\endgroup$
    – chaffdog
    Commented Mar 4, 2018 at 21:34
  • $\begingroup$ It seems that if $\mathcal{C}$ is assumed to be open, then the convexity assumption can be dropped, since ultimately we only care about the directional derivative, in which the direction $l$ can be chosen to have very small length. Then, simply choose $a,b$ to be points sufficiently close so that they are both contained in an open ball $B\Subset\mathcal{C}$, and the ball is clearly convex. $\endgroup$ Commented Dec 1, 2021 at 15:17

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .