To show this linear operator is bounded I introduce a norm on $C^2[0,1]$, namely $||y||_d := ||y||_\infty + ||y'||_\infty + ||y''||_\infty$, where $||.||_\infty$ is the sup norm over $[0,1]$.
Then to show $L$ is bounded, I can show that $||Ly||_\infty\leq M||y||_d$, for all $y\in C^2[0,1]$, and $0<M<\infty$.
I proceed using the triangle inequality
$$||Ly||_\infty=||r(t)y + q(t)y' + y''||_\infty \leq ||r(t)y||_\infty + ||q(t)y'||_\infty + ||y''||_\infty.$$
Now I want to show that the righthand side can be estimated by a positive real number $M$, but I have trouble understanding the norm $||r(t)y||_\infty$ (and similarly $||q(t)y'||_\infty$).
I know that $||y||_\infty =\sup\{y(t):t\in [0,1]\}.$ But for $||r(t)y||_\infty$ there is already a $t$ in there. So is then $||r(t)y||_\infty = \sup\{r(t)y(s) : s\in[0,1]\} = r(t) ||y||_\infty$?
If it is this interpretation, then I think we have $$||r(t)y||_\infty + ||q(t)y'||_\infty + ||y''||_\infty \leq M_1(t) ||y||_d$$ with $M_1(t):=\max\{r(t),q(t),1\}$, but I do not understand whether or not this has to hold for all $t$. I have trouble understanding this since we're taking a norm of functions, not of these function's variables.
If it has to I have to take the sup norm of $r$ and $q$ in the maximum as well. So $M_2:=\max\{ ||r||_\infty, ||q||_\infty, 1\}$, and since $r,q\in C[0,1]$, they're bounded on that interval and this maximum is not infinite (and also $M_1(t)$ is not infinite for any $t$).
So then we get $||r(t)y||_\infty + ||q(t)y'||_\infty + ||y''||_\infty \leq M_1(t) ||y||_d$ for all $t$, so in particular it is $\leq M_2 ||y||_d$? And thus $L$ is bounded.
For example, \begin{align*} |q(t)y'(t)|&\leq\sup_{t\in[0,1]}|q(t)y'(t)|\\ &\leq\sup_{t\in[0,1]}|q(t)|\cdot\sup_{t\in[0,1]}|y'(t)|\\ &=\|q\|_{\infty}\cdot\|y'\|_{\infty}, \end{align*} there is no more $t$ in the end.