Prove that, if a series of continuous functions converges uniformly, then the sum function is also continuous.
This is an exercise in my textbook, Real Analysis and Foundations by Steven Krantz. Could somebody tell me what the sum function means in the context? Does it intend us to prove that $\displaystyle\sum_{j=1}^{\infty}{f_j} = f$ is continuous? If so, $\forall s$ in the domain, we can simply expand $|f(x) – f(s)| \leq |f(x) – f_j(x)| + |f_j(x) – f_j(s)| + |f_j(s) – f(s)|$. The first and the third can be made less than $\frac{\epsilon}{3}$ by uniform convergence (we don't even need uniform convergence I think; pointwise convergence would suffice), and the middle term can be made less than $\frac{\epsilon}{3}$ by continuity of $f_j$ $\forall j$.
If I am understanding the question correctly, is my proof correct?
Thank you!
Best Answer
Hint: Note that pointwise convergence is not sufficient to guarantee the inequality; the problem is that, given any $\varepsilon > 0$, there may not exist some $N \geq 1$ such that the first and the last term on right-hand side is $< \varepsilon/3$ for all suitable $x$. However, if we have uniform convergence, then, given any $\varepsilon > 0$, there is by definition some $N \geq 1$ such that the first and the last term on right-hand side is $< \varepsilon/3$ for all $j \geq N$ and all suitable $x$, and there is by continuity of $f_{N}$ some $\delta > 0$ such that the second term on the right-hand side is $< \varepsilon/3$ for all $|x-s| < \delta$; hence $|x-s| < \delta$ and $j := N$ imply that the right-hand side is $< \varepsilon$.