Help me formulate an example of a sequence of step functions on $(0,1)$ which does not uniformly converge to a bounded continuous function $f$.

Question

I am having trouble coming up with such an example.

How do I go about starting this?

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EDIT:

I have asked the wrong question. Please refer to this question instead:

Formulate an example of a bounded continuous function on $(0,1)$ for which there does not exist a sequence of step functions which uniformly converge.

Answer 1

Let $f(x) = 0$ on $(0,1)$ and $f_n(x) = (1- 1/n)^n 1_{[1- 1/n, 1)}.$

We have $f_n(x) \to 0 = f(x)$ pointwise but not uniformly since $(1 - 1/n)^n \to e^{-1}.$

With fixed $x \in (0,1)$ there exists $N \in \mathbb{N}$ such that $x < 1 - 1/N$. Thus for any $\epsilon > 0$ we have $|f_n(x ) - f(x)| = |f_n(x)| < \epsilon$ for all $n > N$ trivially since $f_n(x) = 0$ for such $n$.

However, $\lim_{n \to \infty}\sup_{x \in (0,1)}|f_n(x)-f(x)| = \lim_{n \to \infty}(1-1/n)^n = e^{-1} \neq 0.$