Uniform convergence on interval.

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Let $a,b$ be real numbers.

There is given a sequence of functions $(f_{n})_{n\ge 1}$. Where $f_{n}:[a,b]\rightarrow \mathbb{R}$ and these functions are smooth. The same with $f:[a,b]\rightarrow\mathbb{R}$

Is it true that:

If this sequence converges pointwise to function $f$, then also this sequence converges uniformly to $f$ ?

I think this is true because of the fact that these functions are bounded. Nevertheless i hope for your help.

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No. Take$$\begin{array}{rccc}f_n\colon&[0,1]&\longrightarrow&\mathbb R\\&x&\mapsto&nx^n(1-x).\end{array}$$Each $f_n$ is a smooth function and $(f_n)_{n\in\mathbb N}$ converges pointwise to the null function. But the convergence is not uniform, since$$(\forall n\in\mathbb N):f_n\left(\frac n{n+1}\right)=\left(\frac n{n+1}\right)^{n+1}$$and $\lim_{n\to\infty}\left(\frac n{n+1}\right)^{n+1}=e^{-1}$.