Let $f_n:\Bbb R\to \Bbb R$ be a continuously differentiable sequence of functions and assume that the sequence $f_n^{'}$ converges uniformly on $\Bbb R$. Also Assume that the sequence $f_n(0)$ converges.

Prove that the sequence is pointwise convergent.

My try:

Assume that $f_n^{'}(x)$ converges uniformly to $g(x)$.

So forall $x\in \Bbb R$ there exists $m\in \Bbb N$ such that forall $n\ge m$ ,$|f_n^{'}(x)-g(x)|<\epsilon$ where $\epsilon>0$ is arbitrary.

I take $f(x)=\int_0^x g(y) dy$

I claim that $f_n(x)$ converges pointwise to $f(x)$.

Also my claim means that $f_n(0)\to 0$ pointwise.

I dont understand if I am right or wrong?

If my claim is correct then how should I prove it?

Can someone help me please?


1 Answer 1


It makes no sense to say that the sequence $\bigl(f_n(0)\bigr)_{n\in\mathbb N}$ converges pointwise to $0$; it is a numerical sequence, not a sequence of functions.

But your idea of defining $f(x)$ as $\int_0^xg(t)\,\mathrm dt$ is a good one. However, you should define it as $\lim_{n\to\infty}f_n(0)+\int_0^xg(t)\,\mathrm dt$. Then, for each $x\in\mathbb R$,\begin{align}f(x)&=\lim_{n\to\infty}f_n(0)+\int_0^xg(t)\,\mathrm dt\\&=\lim_{n\to\infty}f_n(0)+\int_0^t\lim_{n\to\infty}f_n'(t)\,\mathrm dt\\&=\lim_{n\to\infty}f_n(0)+\lim_{n\to\infty}\int_0^xf_n'(t)\,\mathrm dt\text{ (because the convergence is uniform)}\\&=\lim_{n\to\infty}f_n(0)+\lim_{n\to\infty}\bigl(f_n(x)-f_n(0)\bigr)\\&=\lim_{n\to\infty}f_n(0)+\lim_{n\to\infty}f_n(x)-\lim_{n\to\infty}f_n(0)\\&=\lim_{n\to\infty}f_n(x).\end{align}


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.