# Pointwise Convergence to 0 Implies Uniform Convergence to 0

I have seen some related posts on Dini's Theorem, and am actually working a problem related to it, but I have come across some troubling logic unrelated to the theorem. I believe my question to be deserving of its own post. I'll begin by presenting the definitions that have gotten me tangled up. The following definition was given in Ross' Elementary Analysis:

A sequence of real-valued function $f_n \rightarrow f$ pointwise on $S$ means exactly the following $$\text{for each } \epsilon > 0 \text{ and } x \text{ in } S \text{ there exists } N \text{ such that } \\ \left|\;f_n(x)- f(x)\right | < \epsilon \text{ for } n > N \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(1)$$

Later he provides the following condition as equivalent to the $\epsilon-\delta$ formulation of uniform convergence on $S$ $$\textit{A sequence } f_n \textit{ of functions on a set } S \subseteq \mathbb{R} \textit{ converges uniformly to a function } f \textit{ if and only if }\\ \lim_{n \rightarrow \infty} \sup \{ \; \left| \; f(x)-f_n(x) \right| : x \in S \} = 0 \;\;\;\;\;\;\;\;\;(2)$$ Here is my question:

Doesn't $(1) \implies (2)$ in the case that $f_n \rightarrow f$ pointwise on $S$, where $f(x) = 0$ for all $x \in S$?

First off, let me say that I understand that uniform convergence is a much stronger trait than pointwise convergence, and that this is actually the source of my confusion; uniform convergence shouldn't be equivalent to pointwise convergence in this case. Furthermore, I've seen Ross' examples where $f_n \rightarrow f$ pointwise where $f =0$ on all of $S$, but not uniformly, so I understand that there is a fault in my logic, but I can't seem to find it. I'll present my argument now and hopefully get this clarified.

Take any $x_0 \in S$, $\epsilon > 0$ and we have that $f_n(x_0) \rightarrow 0$. Thus by (1) we have $$\exists N_0: n > N_0 \implies \left| \, f_n(x_0) \, \right| < \frac{\epsilon}{2}$$ Since $x$ is arbitrary, this yields $$\exists N: n > N \implies \{ \, \left| \, f_n(x) \, \right| : x \in S \} < \frac{\epsilon}{2}$$ Now we have $$\exists N: n > N \implies \sup \{ \, \left| \, f_n(x) \, \right| : x \in S \} \leq \frac{\epsilon}{2} < \epsilon \;\;\;\;\;\;(3)$$ (3) is equivalent to (2) so the claim holds.

So what's the problem here? What am I doing wrong? I am really lost here, as I truly can't find any errors in the above logic. Any help is appreciated.

• I am confused. Isn't (1) point wise convergence? to see that point wise and uniform convergence isn't the same, consider $f_n(x) = x^n$ on $(0,1)$. – user251257 Aug 5 '15 at 15:54
• @P7E: Your error comes from the fact that your $N$ depend of $x$... – Tryss Aug 5 '15 at 15:55

## 1 Answer

Your $x$ is arbitrary. But $N_0$ depends on $x$. What if $N_0$ gets larger and larger as, say, you get close to the boundary of $S$? Consider $f_n(x)=x^2/n$. Clearly $f_n\to 0$ pointwise. But the $N_0$ such that $|f_n(x)-f(x)|<\epsilon$ for $n>N_0$ depends on $x$. In particular, the larger is $x$, the larger $N_0$ has to be. So you cannot find a value for $N_0$ that "works for every $x$".

Hope this helps.