Does $\Vert f-s_n \Vert_\infty \to 0$ still hold for $f\in C^0[a,b]$?

Question

If $f\in C^2[a,b]$ and $s_n$ its piecewise linear interpolation at points $x_0, \ldots, x_n$ with $h_n = \max_{j=0,\ldots,n-1} (x_{j+1}-x_j)$ then one can show that $$\Vert f-s_n \Vert_\infty \leq \frac{1}{8} h_n^2 \Vert f'' \Vert_\infty. $$

So $h_n \to 0$ implies $\Vert f-s_n \Vert_\infty \to 0$.

But would it still hold if $f\in C^0[a,b]$? I'm pretty sure that yes, yet I have problems showing it.

It's easy to show the statement for $\Vert \cdot \Vert_2$ yet it doesn't help since $\Vert \cdot \Vert_2 \leq (b-a) \Vert \cdot \Vert_\infty$ but there is no bound the other way around.

Any hints are hugely appreciated.

Answer 1

$f\in C^0[a,b]$ is uniformly continuous on $[a, b]$: For every $\varepsilon > 0$ there is a $\delta > 0$ such that $|x-y| < \delta$ implies $|f(x) - f(y)| < \varepsilon$.

Then $h_n < \delta$ implies $\Vert f-s_n \Vert_\infty < \varepsilon $.