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]?

  • 3
    $\begingroup$ Hint: equicontinuity and arzela-ascoli. $\endgroup$ – user251257 Mar 28 '16 at 5:30

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 $[a,b].$

  • $\begingroup$ 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. $\endgroup$ – 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 $\lim_{n\to \infty}(x-x_n)=0.$

  • $\begingroup$ Looks good. The existence of $T$ is because of Bolzano Weirstrass, correct. $\endgroup$ – 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 $n\ge n_0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.