Identity map from sup norm to $L^2$

Question

I'm struggling with the apparently simple task to show that the identity map $Id:(C[a,b],||\cdot||_\infty)\rightarrow (C[a,b],||\cdot||_2)$ is continuous.
Taking $x\in C[a,b]$, I'm trying to find a $\delta$ so that: $$||x-y||_\infty = \max|x(t)-y(t)|<\delta \implies$$ $$||x-y||_2<\epsilon \iff \sqrt{\int_a^b(x(t)-y(t))^2dt}<\epsilon$$ Expanding: $$\sqrt{\int_a^b x^2(t)dt-2\int_a^b x(t)y(t)dt+\int_a^b y^2(t)dt}<\epsilon$$ But then I'm stuck...
My idea:
$\max|x(t)-y(t)|<\delta \implies \int |x-y|<\delta(b-a)\implies\int x<\delta(b-a)+\int y$
But how does that relate to $\int x^2$?

Answer 1

You were very much on the right track!

$$\begin{align}\max|x(t)-y(t)|<\delta &\implies \max|x(t)-y(t)|^2<\delta^2 \\ &\implies \int_a^b |x-y|^2<\delta^2(b-a)\\ &\implies\sqrt{\int_a^b |x-y|^2}<\delta \sqrt{b-a} \end{align}$$ So, $\|x - y\|_2 < \delta \sqrt{b-a}$, thus choose $\delta = \frac{\epsilon}{\sqrt{b-a}}$.

Answer 2

The first thing you notice in your idea is the right observation but the trick is to apply it sooner since expanding $(x(t)-y(t))^2$ only makes things harder for you!

Instead, if $\max |x(t) - y(t)| < \delta$ then $$\int_a^b (x(t)-y(t))^2 dt \leq \int_a^b \delta^2 dt = \delta^2 (b-a).$$

From here, you just need to play around to pick the right value of $\delta$.