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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?

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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}

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