Find a function that has a 1 or 2 norm but not an infinity norm.

Question

In the lecture, the lecturer told us that there exist functions that have a 1-norm or a 2-norm but no infinity-norm. Or to be more formal:

1. What are examples of functions satisfying $f \in \mathcal{L}_1$, $f \not\in \mathcal{L}_{\infty}$.
2. Secondly what is an example of a function satisfying $f \in \mathcal{L}_2$, $f \not\in \mathcal{L}_{\infty}$.
3. The last thing that I was wondering was that the lecturer also asked if we could come up with a function satisfying $f \in \mathcal{L}_2, \lim_{x\to\infty}f(x)=0$.

edit: For point 3 the function I found was $f(x)=\frac{1}{x+1}$ which satisfies the demand.

Where $\mathcal{L_p}$ is defined as $\left( \int_0^\infty x(t)^p dt \right)^{\frac{1}{p}}$, and in the case of $\mathcal{L_\infty}$ it's defined as $\sup_{t \geq 0}|x(t)|_\infty$.

To me it seems counterintuitive to think of a function with a finite integral, but without a supremum. Where is my thinking going astray? Or does it have to do with the domain that you specify for the function in some form or another?

Answer 1

Roughly speaking, you should try to find a function with a vertical asymptote which is near enough from the graph. Note for example that

$$\int_0^1\frac{dt}{\sqrt{t}}=2$$

Then, if $f(t)=t^{-1/2}$ and $g(t)=t^{-1/4}$, $$\|f\|_1=2$$ and $$\|g\|_2=\sqrt 2$$

Answer 2

Consider $x \mapsto x^{-1/2}$ on $(0,1)$: it's in $L^1$ (with integral $2$), but not bounded above near $0$, so is not $L^{\infty}$.

The other can be done in exactly the same way by replacing $-1/2$ with a different power.