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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.

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  • $\begingroup$ Uniform convergence on any set implies Uniform convergence on any subset by definition. $\endgroup$ – Kavi Rama Murthy Aug 14 at 6:26
  • $\begingroup$ @KaviRamaMurthy, Follow-up question: Prove that $\sum_{n=1}^{\infty}\frac{\sin(nx)}{n}$ is uniformly convergent (or not) on $(-1,1)$. By Dirichlet's test, $\sum_{n=1}^{\infty}\frac{\sin(nx)}{n}$ is uniformly convergent on $[0,1] \subset [0,2\pi]$ Now, being periodic with period $2\pi$, it would be uniformly convergent on $[-1,0] \subset [-2\pi, 0]$. Thereby it is uniformly convergent on $[-1,1]$ and on $(-1,1)$. Is is correct? $\endgroup$ – justanotheruser Aug 14 at 6:41
  • $\begingroup$ This series is not uniformly convergent on $[0,2\pi]$. $\endgroup$ – Kavi Rama Murthy Aug 14 at 6:43
  • $\begingroup$ @KaviRamaMurthy Proof: (That the series is not uniformly convergent on $[0,2\pi]$) For every $n \in \mathbb{N}$, $\frac{1}{2n} \in$ $[0,2\pi]$. Now, $|s_{2n}(\frac{1}{2n})-s_{n}(\frac{1}{2n})|=|\frac{\sin(\frac{n+1}{2n})}{n+1}+\frac{\sin(\frac{n+2}{2n})}{n+2}+...+\frac{\sin(\frac{2n}{2n})}{2n}|\geq \sin(\frac{1}{2})|\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}| \geq \sin(\frac{1}{2})|\frac{1}{2n}+...+\frac{1}{2n}|\geq \frac{\sin(\frac{1}{2})}{2}= \epsilon_0$ $s_n(x)$ being the partial sum function. Is is correct now? (Cauchy Criterion for uniform convergence is not satisfied) $\endgroup$ – justanotheruser Aug 14 at 7:51
  • $\begingroup$ Looks fine to me. There is also a theorem that if $a_n$ decreases to $0$ then $\sum a_n \sin (nx)$ converges uniformly iff $na_n \to 0$. Ref: Fourier Series by Edwards $\endgroup$ – Kavi Rama Murthy Aug 14 at 7:55
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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 $n\geq k$.

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