Prove that if $f_n$ converges pointwise and $|f_n'| < M$, then $f_n$ converges uniformly

Suppose $f_n(x)$ is a sequence of functions with continuous derivatives in $[a, b]$. How do I prove that if $f_n(x)$ converges pointwise and $|f_n'(x)| < M$ for all $n$ and $x \in [a,b]$ (where $M$ is a constant), then the convergence is uniform on $[a,b]? • Hint: equicontinuity and arzela-ascoli. – user251257 Mar 28 '16 at 5:30 3 Answers Given$x,y \in [a,b]$, by the MVT there exists$\xi_n$between$x$and$y$such that $$|f_n(x) - f_n(y)| = |f_n'(\xi_n)(x-y)| \leqslant M|x-y|.$$ Since$f_n \to f$pointwise, $$|f(x) -f(y)| = \lim_{n \to \infty}|f_n(x) - f_n(y)|\leqslant M|x-y|.$$ Let$\epsilon > 0$be given and set$\delta = \epsilon/(3M)$. Since$[a,b]$is compact there exists a finite number of points$c_1, \ldots, c_m$such that $$[a,b] \subset \bigcup_{j=1}^m(c_j - \delta,c_j+\delta).$$ Again by pointwise convergence, there exists$N_j \in \mathbb{N}$such that if$n > N_j$then $$|f_n(c_j) - f(c_j)| \leqslant \epsilon/3.$$ Let$N = \max_{1 \leqslant j \leqslant m}N_j$. For any$x \in [a,b]$there exists$k \in \{1,2,\ldots,m\}$such that$x \in (c_k - \delta,c_k + \delta).$We have $$|f_n(x) - f(x)| \leqslant |f_n(x)- f_n(c_k)| + |f_n(c_k) - f(c_k)| + |f(c_k) - f(x)|$$. If$n > N \geqslant N_k$, then since$|x - c_k| < \delta = \epsilon/(3M)$, we have $$|f_n(x) - f(x)| < M \delta + \epsilon/3 + M \delta = \epsilon.$$ Therefore,$f_n$converges uniformly to$f$on$[a,b].$• Nicely done. I'm once again having my text displayed "as typed" ($ and all that) unless I exit the Q and re-enter, which makes editing a pain. – DanielWainfleet Mar 28 '16 at 7:29

Let $f(x)=\lim_{n\to \infty}f_n(x)$ for $x\in [a,b].$

By contradiction, suppose $(f_n)_{n\in N}$ does not converge uniformly. Then for some $r>0,$ there exists an infinite $S \subset N$ and a sequence $(x_n)_{n\in S}$ of members of $[a,b]$ such that $\forall n\in S\; (|f_n(x_n)-f(x_n)|>r.$ Now there exists an infinite $T\subset S$ such that $(x_n)_{n\in T}$ converges to a limit $x$.

Now for $n\in T$ we have $r<|f_n(x_n)-f(x_n)|\leq |f_n(x_n)-f_n(x)|+|f_n(x)-f(x_n)|$. But $|f_n(x_n)-f_n(x)|= |\int_{x_n}^x f'_n(z)\;dz|\leq M |x_n-x|,$ which tends to $0$ as $n$, in $T,$ ,goes to $\infty.$ So there are infinitely many $n\in T$ for which $$r/2<|f_n(x)-f(x_n)|.$$ But $f_n(x)\to f(x)$ as $n\to \infty$, so there are infinitely many $n\in T$ such that $$r/4<|f(x)-f(x_n)|.$$

Now for each such $n,$ we have $|f(x)-f(x_n)|=$ $\lim_{m \to \infty}|f_m(x)-f_m(x_n)|\leq$ $\lim \sup_{m\to \infty}\int_{x_n}^x|f'_{m}(z)|\;dz \leq$ $M|x-x_n|$ which means that $r/4<M|x-x_n|$ for infinitely many $n\in T.$

This is absurd because $r>0$ and $\lim_{n\to \infty}(x-x_n)=0.$

• Looks good. The existence of $T$ is because of Bolzano Weirstrass, correct. – RRL Mar 28 '16 at 7:43

Step 1. The $f_n$'s and $f$ are Lipschitz continuous in $[a,b]$ with constant $M$.

Indeed, by virtue of the Mean Value Theorem, $\,f_n(x)-f_n(y)=f'_n(\xi)(x-y)$, for some $\xi\in(x,y)$, and hence $$\lvert \,f_n(x)-f_n(y)\rvert =\lvert\,f'_n(\xi)\rvert\lvert x-y\rvert\le M \lvert x-y\rvert,$$ and pointwise convergence of $\{f_n\}$, and in particular of the sequences $\{f_n(x)\}$ and $\{f_n(y)\}$ provides that $$\lvert \,f(x)-f(y)\rvert=\lim_{n\to\infty}\lvert \,f_n(x)-f_n(y)\rvert\le M \lvert x-y\rvert.$$

Step 2. $f_n\to f$ uniformly.

Let $\varepsilon>0$ and pick $$a=y_0<y_1<\cdots<y_m=b,$$ so that $\lvert y_j-y_{j-1}\rvert<\frac{\varepsilon}{3M}$, and hence $$\lvert f_n(y_j)-f_n(y_{j-1})\rvert<\frac{\varepsilon}{3},\,\,n\in\mathbb N, \,\,\,\text{and}\,\,\, \lvert f(y_j)-f(y_{j-1})\rvert<\frac{\varepsilon}{3}.$$ Let now an arbitrary $x\in[a,b]$. Then $x\in [y_{j-1},y_j]$, for some $j=1,\ldots,m$. Let $n_0\in\mathbb N$, so that $$\lvert\, f_n(y_j)-f(y_{j})\rvert<\frac{\varepsilon}{3},\quad j=0,\ldots,m, \quad\text{for all n\ge n_0}.$$ Such $n_0$ exists, due to the pointwise convergence of $\{f_n\}$. Then for an arbitrary $x\in[a,b]$, we have $$\lvert\, f_n(x)-f(x)\rvert\le\lvert\, f_n(x)-f_n(y_j)\rvert +\lvert\, f_n(y_j)-f(y_j)\rvert++\lvert\, f(y_j)-f(x)\rvert < \frac{\varepsilon}{3}+\frac{\varepsilon}{3}+\frac{\varepsilon}{3}=\varepsilon,$$ for all $n\ge n_0$.