# $f_n$ converge uniformly to $f$ then $\mathrm{d}f_n(x_n)$ converges to $\mathrm{d}f(x)$

Let $$f_n : \mathbb{R}^p \to \mathbb{R}$$ such that the $$f_n$$ are $$C^1$$ and such that the sequence $$(f_n)_{n \in \mathbb{N}}$$ converges uniformly to a function $$f : \mathbb{R}^p \to \mathbb{R}$$ which is $$C^1$$. Then prove that for all $$x \in \mathbb{R}^n$$ there is a sequence $$(x_n)_{n \in \mathbb{N}}$$ which converge to $$x$$ such that $$\mathrm{d}f_n(x_n)$$ converges to $$\mathrm{d}f(x)$$.

I must say that I don't know at all how to do and don't have any intuition of what is really going on here. So we might look at the case qhere $$p= 1$$.

So we can write :

$$f(a+h) = f(a)+ f'(a)h +o(h)$$ $$\forall n \in \mathbb{N}, f_n(a+h) = f_n(a) + f'_n(a)h +o(h)$$

Hence we have :

$$\mid f'(a)h - f'_n(a)h \mid \leq \mid f(a+h)-f(a) \mid +\mid f(a)-f_n(a) \mid + \mid o(h) \mid$$

Since the function $$f_n$$ converge uniformly to $$f$$, we have : $$\mid f'(a)h - f'_{\infty}(a) \mid \leq \mid o(h) \mid$$ And now using we let $$h \to 0$$ so that :

$$\mid f'(a) -f_\infty'(a) \mid = 0$$

I don't know if this works, but it feels strange to me since in the case the sequence $$x_n$$ is just the constant sequence... and moreover if this is correct I don't see at all how to generalise to higher dimensions.

Thank you !

• Your demonstration is not correct, since you don’t know if $lim_{n \to \infty} f’_n(a)$ converges, hence you can note it like $f’_{\infty}(a)$, that’s why you always need to be careful when dealing with double limit – Thinking Jan 28 '19 at 23:42
• For complex or vector valued functions the claim is wrong. Consider the sequence $f_n(x):={1\over n}e^{inx}$, which converges uniformly to $0$, but $|f_n'(x)|=1$ for all $x$ and $n$. – Christian Blatter Jan 31 '19 at 10:34

We first prove the result in the case where there is a local maximum. At a maximum, the differential vanishes, so the claim is here:

Lemma. If $$(g_n)_n$$ is a sequence of $$C^1$$ functions $$\mathbb{R}^p\to \mathbb{R}$$ converging uniformly to a $$C^1$$ function $$g$$ having a local (strict) maximum at $$y$$, that is $$g(x) for all $$x\neq y$$ in a ball neigborhood $$B(y,r)$$ of $$y$$, then there exist a sequence $$x_n$$ such that $$\lim x_n=y$$ and $$dg_n(x_n)=0$$ for sufficiently large $$n$$.

Proof of the Lemma: Pick $$N$$ sufficiently large so that $$\forall n\geq N$$, $$sup_{ \|x-y\|=r} g_n(x) The existence of such an $$N$$ follows from the fact the corresponding inequality is true for $$g$$ by hypothesis, and the uniform convergence of $$g_n$$. For any $$n\geq N$$, pick $$x_n$$ to be a maximum of $$g_n$$ on $$B(y,r)$$. Because of the previous inequality, $$x_n$$ is in the interior of the ball $$B(y,r)$$. So the derivative satisfies $$dg_n(x_n)=0$$.

Let $$x$$ be a limit point of a subsequence of $$(x_n)$$. Since $$g_n(x_n)\geq g_n(y)$$ by definition of $$x_n$$, taking the limit we have $$g(x)\geq g(y)$$, and of course $$x_n$$ is still in the closed ball $$B(y,r)$$. So necessarily $$x=y$$ since $$y$$ is a local strict maximum of $$g$$ on $$B(y,r)$$. This concludes the proof of the Lemma.

Now, to deal with the general case, pick a point $$y$$ and define $$g_n(x)=f_n(x)-f(x)-\|x-y\|^2.$$ Clearly, this sequence of $$C^1$$ functions converges to $$g(x)=-\|x-y\|^2.$$ We now apply the Lemma, so there is a sequence $$(x_n)_n$$ such that $$\lim x_n=y$$, and $$dg_n(x_n)=0$$. But since $$dg_n(x_n).h=df_n(x_n).h-df(x_n).h-2\langle x_n-y, h \rangle,$$ so $$df_n(x_n)=df(x_n)+ 2\langle x_n-y, . \rangle.$$ The linear forms $$h\mapsto 2\langle x_n-y, h \rangle$$ converges to zero because of Cauchy-Scharwz inequality, and $$df(x_n)$$ converges to $$df(y)$$ since $$f$$ is $$C^1$$. This concludes the proof.

• This is clever ! I am wondering why it's actually simpler to prove the problem when there is a local minima/maxima. I guess it's because in this case we know what is the value $\mathrm{d}f$ at this point. I'll deliver the bounty tonight. Thank you ! – dghkgfzyukz Jan 31 '19 at 18:59

In one dimension we don't need uniform convergence; pointwise converge will do. Also we only need $$f$$ and $$f_n, n=1,2,\dots$$ differentiable everywhere, not necessarily $$C^1.$$

WLOG we can assume $$f\equiv 0$$ because $$f_n(x)-f(x)\to 0$$ everywhere and $$f_n-f$$ is differentiable everywhere.

Fix $$x\in \mathbb R.$$ Let $$\delta > 0.$$ Then for $$n\in \mathbb N$$ the MVT shows there exists $$c(n,\delta)\in (x,x+\delta)$$ such that

$$f_n(x+\delta)- f_n(x) = f_n'(c(n,\delta))\delta.$$

Since the left side $$\to 0$$ as $$n\to \infty,$$ we can make the right side as small as we like by taking $$n$$ large. We can thus find $$N = N_\delta$$ such that $$n\ge N_\delta$$ implies $$|f_n'(c(n,\delta))| < \delta.$$

Now think of $$\delta_k = 1/k, k=1,2,\dots .$$ Then from the above there exist integers $$0 such that $$N_k\le n < N_{k+1}$$ implies $$|f_n'(c(n,1/k))| < 1/k.$$ If we then define

$$x_n = c(n,1/k),\,\, N_k\le n N_{k+1},$$

we have $$x_n\to 0$$ and $$f'(x_n)\to 0.$$ (We can let $$x_n$$ be anything for $$1\le n )

• You can't say WLOG $f = 0$ since at the beginning you are assuming $f$ is not necessarily $C^1$. – Thinking Feb 5 '19 at 19:02
• @Thinking I don't understand your comment. – zhw. Feb 5 '19 at 21:45
• It's not so clear for me that you are not loosing generality by saying $f = 0$, since $f$ is not necessarily $C^1$. – Thinking Feb 5 '19 at 22:18
• @Thinking Like I wrote, I am assuming $f$ and $f_n, n=1,2,\dots$ are differentiable everywhere. Where do you think the proof goes wrong? – zhw. Feb 5 '19 at 22:22