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  • $(f_n)$ converges uniformly on $(a , b)$
  • $(f_n)$ converges pointwise at $x = a$ and $x = b$

Prove that $(f_n)$ converges uniformly on $[a , b]$.

I am confused on how to show this, specifically at the endpoints. Using Cauchy's criterion and looking at $x = a$, we can write \begin{align*} |f_n(a) - f_m(a)| &= |f_n(a) - f_n(x) + f_n(x) - f_m(x) + f_m(x) - f_m(a)| \\ &\leq |f_n(a) - f_n(x)| + |f_n(x) - f_m(x)| + |f_m(x) - f_m(a)| \end{align*}

but I am lost after that. Maybe that is not the way to go about it.

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    do you have continuity of $f_n , f$? – 2017-01-08
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    Unfortunately, no – 2017-01-08
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    There is no need to think for continuity here. Just definition of uniform convergence is sufficient. – 2017-01-08

1 Answers 1

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Given that $(f_n)$ converges uniformly on $(a,b)$ to say $f$, we know that $\forall \epsilon>0$ $\exists \ N_0\in \mathbb{N}$ such that $\forall x\in(a,b)$ $|f_n(x)-f(x)|<\epsilon$.

Also, $(f_n)$ converges pointwise to $f$ at $a$ and $b$, hence $\exists$ $N_1$ and $N_2\in \mathbb{N}$ such that $\forall \epsilon>0$ $|f_n(a)-f(a)|<\epsilon$ and $|f_n(b)-f(b)|<\epsilon$ (respectively).

Choose $N=max\{N_0,N_1,N_2\}$ and see that $(f_n)$ converges uniformly on $[a,b]$.

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    Ok, that makes sense. Thanks! – 2017-01-08