Converge pointwise but not uniformly a problem from Axler 2020

Question

I am studying Real analysis using the book of Axler (Measure Integration and Real analysis) in which the author gives an example about a function that converges pointwise but not uniformly as follows:

However, it seems to me that it is possible to prove that $f_k$ converges uniformly to $f$ as follows:

+) If $x \neq 0$
Given $\epsilon > 0$, we have $|f_k(x) - f(x)| = |f_k(x) - 1|$
For every $x \neq 0$, there exists a natural number $m$ such that $m > \frac{1}{x}$ (i.e. $x \gt \frac{1}{m}$).
With $k \geq m$, $f_k(x) = 1$ and so $|f_k(x) - f(x)| = |f_k(x) - 1| = |1-1| = 0 \lt \epsilon$ for all $k \geq m$

+) If $x = 0$, given $\epsilon > 0$, we have |$f_k(0) - f(0)$| = $0 \lt \epsilon$ for all $k$
So that for all $k \geq m$ ($m$ is taken from above), we have |$f_k(0) - f(0)$| = $0 \lt \epsilon$

To conclude, for every $\epsilon \gt 0$, there exists a natural number $m$ such that $|f_k(x) - f(x)| \lt \epsilon$ for all $k \geq m$ and for all $x \in X$.

As for me, the proof makes sense which contradict the claim of the author. Could you please help me explain this issue ?

Answer 1

Remember that uniform convergence preserves continuity. In particular, consider a sequence $\{f_n\}$ of continuous functions, where $f_n: [a,b] \to \mathbb{R}$. Then, $f_n \to f$ uniformly on $[a,b]$ implies that $f$ is continuous.

The problem with your proof is that you can't pick one value of $m$ that will work for all values of $x$.

Answer 2

Uniformly means you need the same $m$ for all $x.$ You specifically give $m$ in terms of $x.$

You’ve just shown that the sequence converges pointwise, not uniformly.

You can prove it doesn’t converge uniformly by choosing $\epsilon<\frac12.$ Then, given an $m,$ you can pick $x=\frac1{2m}.$ Then $f_m(x)=\frac{3}2$, so: $$\left|f_m\left(\frac1{2m}\right)-f\left(\frac1{2m}\right)\right|=\frac12>\epsilon$$