Is this approach for this particular question correct

Question

I have been trying some questions in analysis and this was one of the questions asked .

Let ${f_{n}}$ be a sequence of continuous functions on [0,1] such that $\sum_{n=1}^{\infty}f_{n} $ converges uniformlly on (0,1] .Show that $\sum_{n=1}^{\infty}f_{n}(0)$ converges .

I thought of using the result : If $f_{n}$ converges to f uniformly and and each $f_{n}$ is continuous at c then limit function is also continuous at c .

By this result I can easily deduced what is asked . Am i right ?

If I am wrong then kindly tell me the reason and also the method which should be used .

Thank you !!

Answer 1

I'll suggest an $\frac{\epsilon}3$ argument. Let $s_N(x)=\sum_{n=1}^N f_n(x)$ and $s(x)=\lim_{N\rightarrow\infty}s_N(x)$. You know that $s(x)$ exists and is continuous on $(0, 1]$. Morover the $f_n(x)$ are continuous on a compact set $[0, 1]$, so their are uniformly continuous on it and, in particular, on the subset $(0, 1]$. That property extends to $s_n(x)$ and thus $s(x)$ is also uniformly continuous as the limit of a sequence of uniformly continuous function. Hence $s(x)$ can be uniquely extended to an uniformly continuous function on $[0,1]$.

We will show that $s(0)=\lim_{N \rightarrow \infty}s_N(0)$. Firstly, using the fact that $\{s_N(x)\}$ converges uniformly, we can find an $M$ such that $N\ge M \implies \left| s(x) - s_N(x) \right|< \frac{\epsilon}3$ for every $x\in(0, 1]$. Secondly, due to the fact that $s_N(x)$ is uniformly continuous, we can find a $\delta_1$ such that $\left| 0 - y \right| < \delta_1 \implies \left| s_N(0) - s_N(y) \right| < \frac{\epsilon}3$. Finally, using the fact that $s(x)$ is uniformly continuous, we can find a $\delta_2$ such that $\left| 0 - y \right| < \delta_2 \implies \left| s(0) - s(y) \right| < \frac{\epsilon}3$.

For every $N\geq M$ we can choose a $\delta_1$ and an $y$ such that $\left| 0 - y \right| < \delta=\min\{\delta_1, \delta_2 \}$, so we have $\left| s(0)-s_N\right(0)|\leq \left| s(0)-s\right(y)| + \left|s(y) -s_N(y)\right| + \left| s_N(y)-s_N\right(0)| \lt \epsilon$. Hence $s(0)$ is the limit of $s_N(0)$.

Answer 2

Note that since the convergence is uniform on $(0, 1]$, given $\epsilon > 0$, there is some $N>0$ such that $n > m > N \implies |\sum_{k=m}^n f_k(x)| < \epsilon$, for all $x \in (0, 1]$.

Since each $f_k$ is continuous on $[0, 1]$, letting $x \to 0$, we have $n > m > N \implies |\sum_{k=m}^n f_k(0)| < \epsilon$.

Then the sequence $(\sum_{k=1}^n f_k(0))_{n\ge1}$ is Cauchy, which implies $\sum_{k=1}^\infty f_k(0)$ is convergent.