Vtyjkyuk

Prove that if $f_n$ converges uniformly on $(a,b)$, $f_n(a)$ and $f_n(b)$ converge pointwise. I want to show that $f_n$ converges uniformly on $[a,b]$

MY TRIAL:

Let $epsilon>0$, since $f_n$ converges uniformly on $(a,b)$, then $exists,N_1=N(epsilon)$ s.t. $forall ,ngeq N$, $forall xin (a,b)$
beginalignleft|f_n(x)-f(x)right|<epsilonendalign

Since $f_n$ converges pointwise when $x=a$, then $exists,N_2=N(epsilon,a)$ s.t. $forall ,ngeq N$,

beginalignleft|f_n(a)-f(a)right|<epsilonendalign

Also, $f_n$ converges pointwise when $x=b$, then $exists,N_3=N(epsilon,b)$ s.t. $forall ,ngeq N$,

beginalignleft|f_n(b)-f(b)right|<epsilonendalign

Hence, $forall ,ngeq maxN_1,N_2,N_3,$

$$limlimits_nto inftyf_n(x)=begincasesf(a) & x=a,\f(x)&xin(a,b),\f(b)&x=bendcases$$
Hence, $f_n$ converges uniformly on $[a,b]$.

Please, I'm I right? If no, an alternative proof will be highly regarded. Thanks!

Your proof is correct! In general you can prove that if a sequence of functions $f_n_n$ converges uniformly on a finite set of domains $A_1,dots,A_k$ (in your case the domains are $A_1=a,A_2=(a,b),A_3=b$), then it converges uniformly in its union $cup_k A_k.$

The proof works exactly like yours.

Indeed, for every $epsilon>0$ there exist $N_1,dots,N_k$ such that for every $ngeq N_i$ it is true that $|f(x)-f_n(x)|<epsilon$ for every $xin A_i$.

It is then sufficient to consider $N=max N_1,dots,N_k$ to say that $|f(x)-f_n(x)|<epsilon$ for every $xin cup_i A_i$ and for every $ngeq N,$ which is exactly the concept of uniform convergence.