which norm of these is greater?

Question

I have a vector valued function $x:[a, b]\rightarrow \mathbb{R}^n$ and it is given that $x\in L^2([a,b]:\mathbb{R}^n)$. Then I wanted to know which of these norms have greater value? $||x||_{L^2}=\int_{a}^{b}||x(t)||_{\mathbb{R}^n}^2 \hspace{0.2 cm}dt$ and $||x||_{sup}=sup\{||x(t)||:t\in [a, b]\}$

Answer 1

First, you are missing a square root: $$\|x\|_2=\sqrt{\int_{a}^{b}\|x(t)\|^2\ dt}$$

Since for all $t \in [a, b], \|x(t)\| \le \|x\|_\infty = \sup\{\|x(t)\| : t\in[a,b]\}$, we have that $$\int_{a}^{b}\|x(t)\|^2\ dt \le \int_{a}^{b}\|x\|_\infty^2\ dt = (b-a)\|x\|_\infty^2$$ and so $$\|x\|_2 \le (\sqrt{b-a})\|x\|_\infty$$

This inequality cannot be improved upon, as it is an equality when $\|x(t)\|$ is constant. So the question of which is larger depends on how far away $a$ and $b$ are from each other. When $b - a \ge 1$, then the supremum norm is always larger. when $b - a < 1$, then either norm can be larger, depending on the map $x$.

For a more general comparison, see Hölder's inequality.