# Uniform Convergence of derivative

Question

Suppose a sequence of derivatives of functions $f'_n$ converge uniformly to $f'$ where $f_n$ is defined on the on the interval $[a,b]$. And $f_n(x)$ converges pointwise to $f(x)$ for $x\in \mathbb{R}$. Prove that $f_n$ converges uniformly to $f$.

Effort

I basically just have the definition:

$1.$ $\forall \epsilon >0, \forall n >N, |f'_n(a)-f'(a)|<\epsilon, \forall a\in \mathbb{R}$

and $\forall \epsilon ' >0, \forall n' >N', |f_n(x)-f(x)|<\epsilon '$ for $x\in \mathbb{R}$

$2.$ Another point to note may be is : there exists a point $c$ such that $$\frac{f'_n(c)}{f'(c)}=\frac{f_n(b)-f_n(a)}{f(b)-f(a)}$$ Applying limit to both side$$\lim_{n\rightarrow\infty}\frac{f_n(b)-f_n(a)}{f(b)-f(a)}=1$$

It seems second one is right track but I don't seem to be able to complete it.

P.S. Not sure if it's related to functional analysis.

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Approach 2 might go somewhere if you set $a=x_0$ and $b=x_0+h$.Then you get that $f_n(x_0+h)-f_n(x_0)\to f(x_0+h)-f(x_0)$ uniformly with respect to $h$. Since $f_n(x_0)\to f(x_0)$, you might conclude from this that $f_n(x_0+h)\to f(x_0+h)$ uniformly in $h$, which is the same thing as to say that $f_n(x)\to f(x)$ uniformly in $x$. – Giuseppe Negro Jan 27 '13 at 17:40
Thank You. This was one I was looking for. The integration one is good too. – user45099 Jan 27 '13 at 17:46

A quick approach is via integration. Fix $x_0\in [a, b]$ such that $f_n(x_0)$ converges. Then write $$f_n(x)=f_n(x_0)+\int_{x_0}^x f'_n(y)\, dy.$$ The right hand side converges uniformly to $f(x_0)+\int_{x_0}^x f'(y)\, dy$, because $$\begin{split} \left\lvert f_n(x_0)-f(x_0)+\int_{x_0}^x f'_n(y)-f'(y)\, dy\right\rvert &\le \lvert f_n(x_0)-f(x_0)\rvert +\left\lvert \int_{x_0}^x \lvert f'_n(y)-f'(y)\rvert\, dy\right\rvert \\ &\le \lvert f_n(x_0)-f(x_0)\rvert+(b-a)\lVert f'_n-f'\rVert_\infty, \end{split}$$ and the right hand side is now a numerical sequence which tends to 0 by assumption. Since $$f(x_0)+\int_{x_0}^x f'(y)\, dy=f(x),$$ we have thus proven that $f_n \to f$ uniformly.
The only small disadvantage of this proof is that it requires the fundamental theorem of calculus and so it requires integrability of $f'_n$. This is certainly the case if $f_n\in C^1([a,b])$, but the theorem still holds for more pathological functions which are not the integral of their derivative. For this (small) generalization you can consult Rudin, Principles of mathematical analysis, chapter 7.