Let $u\in C([a,b];X)$ and $(u_n)$ be a sequence of step functions. For all $n \in \mathbb N$ define $$u_n(t):=u(t_{(k-1)}^{(n)})$$ for $t \in [t_{k-1}^{(n)},t_k^n)$, $k=1,2...,2^n$ and $$u_n(b):=u(t_{2^n-1}^{(n)})$$ where $t_k^{(n)}= a+k \frac{b-a}{2^n}$.
How can I show that $(u_n)$ converges uniformly to $u$ or what is the approach?
Some little help is appreciated.
I claim X to be a metric space,then u is uniformly continous so for a given $\varepsilon > 0$ exists a $\delta > 0$ etc…
$u_n$ converges uniformly to $u$ iff $$\forall\varepsilon>0\exists N\in\mathbb{N}\forall n\ge N: ||u_n - u||_\infty < \varepsilon$$
Take such an $\varepsilon$ and choose N that way, that $\frac{b-a}{2^N} < \delta$ then for $n\ge N$ $$\left|u(t_{(k-1)}^{(n)}) - u(t)\right| < \varepsilon$$ for $t \in [t_{k-1}^{(n)},t_k^n) $ so $||u_n - u||_\infty < \varepsilon$