Let us consider a sequence of functions $\{f_n\}$ on a compact interval $[a,b]$, which is uniformly convergent (to a function, say $f$) on $[a,b]$. Does it ensure the uniform convergence of $\{f_n\}$ on the open interval $(a,b)$ to the same function $f$ restricted on $(a,b)$?
(or uniformly convergent to some other function)
I don't know whether or not it is true for sure. It might be true (I think) because:
By uniform convergence of $\{f_n\}$ on $[a,b]$:
$ \forall \; \epsilon>0$, $\exists \; k \in \mathbb{N}$ such that $|f_n(x)-f(x)|_{\forall \; x \in [a,b]}<\epsilon$, $\forall n \geq k$.
Now, for that very same $\epsilon$ and corresponding $k$ we can do:
$ \forall \; \epsilon>0$, $\exists \; k \in \mathbb{N}$ such that $|f_n(x)-f(x)|_{\forall \; x \in (a,b)}<\epsilon$, $\forall n \geq k$.
[The inequality being valid on $[a,b]$, we can infer that it also holds for $(a,b) \subset [a,b]$. ]
Is my argument true? Kindly Verify.
Best Answer
Yes this is true, as you said, for any $\epsilon>0$, you can simply choose the same $k$ to ensure $|f_{n}(x)-f(x)|<\epsilon$ for all $x\in(a,b)$ and $n\geq k$.