Where is the mistake in my counter-example to the statement? (sequences of functions, uniform convergence)

Question

Statement:
Let $f,f_n:[a,b]\to\mathbb{R}$ for $n\in\mathbb{N}$ and for all $x\in[a,b]$ let $\lim_{n\to\infty}f_n(x) = f(x)$.
If $f$ and all $f_n$ are continuous, then $f_n$ converges uniformly against $f$.

Counter example is supposed to be the function $$f_n( x) ≔ \begin{cases} g(n·x) &, 0<x<\frac 1 n\\ 0 &, \text{else} \end{cases}$$

Where $g(x)≔ 8·x^4 - 16·x^3 + 8·x^2$ .

Here is a plot of it for $n$ increasing:

With this function, all requirements of the statement should be fullfilled:

We have $f(x) = \lim_{n\to\infty} f_n(x) = 0$, so $f$ is continuous.

Further, all $f_n$ are continuous, as $\lim_{x\to 0} f_n(x) = 0 = \lim_{x\to \frac 1 n} f_n(x) $.

However, $f_n$ doesn't converge uniformly against $f$, as for $\epsilon = \frac 1 2$, no matter how high we choose $n\in\mathbb{N}$, we have $f_n(\frac 1 {2n}) = \frac 1 2$.

Now, I've found two text books claiming the statement to be true, so somewhere I've got to have made an error in my reasoning. I can't find it though: Where is it?

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Both text books are German (and rather old), the statement is however as faithful translated as possible.
First is: Landers Rogge, Nichtstandardanalysis, page 132, exercise 1 (from 1994).
Second is (cited from wikipedia): F. Hausdorff: Grundzüge der Mengenlehre. 1914, Chelsea Publishing Co., New York 1949, Kap. IX, § 4

Answer 1

Theorem 7.9 in Rudin's Principles of Mathematical Analysis states the following:

Suppose $$\lim_{n\to\infty}f_n(x)=f(x) \qquad (x\in E).$$ Put $$M_n=\sup_{x\in E}|f_n(x)-f(x)|.$$ Then $f_n\to f$ uniformly on $E$ if and only if $M_n\to 0$ as $n\to \infty.$

You can see that for your counterexample, $M_n\not\to 0$ as $n\to\infty$. Also note the "if and only if".