# If a sequence of functions $\{f_n\}$ be uniformly convergent on $[a,b]$, would it be uniformly convergent of $(a,b)$?

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.

• Uniform convergence on any set implies Uniform convergence on any subset by definition. Aug 14, 2019 at 6:26
• @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? Aug 14, 2019 at 6:41
• This series is not uniformly convergent on $[0,2\pi]$. Aug 14, 2019 at 6:43
• @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) Aug 14, 2019 at 7:51
• 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 Aug 14, 2019 at 7:55

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 $$n\geq k$$.