$L^{1}$ and $L^{\infty}$ norms are not equivalent on compact sets

Question

Let $B \subset \mathbb{R}^{n}$ be compact and $C(B)$ be the vector space of continuous functions $f: B \rightarrow \mathbb{R}$. Prove that $$ \begin{aligned} \|f\|_{L^{1}} &:=\int_{B}|f(x)| d x \\ \|f\|_{L^{\infty}} &:=\sup \{|f(x)|: x \in B\} \end{aligned} $$ are two norms on $C(A)$ which are not equivalent.

First, can I say that f is a continuous function on a compact set and thus Riemann integrable? and that Riemann integral will be equal to Lebesgue integral? I thought to take an example for 1D but not sure that it's enough to show that these norms are not equivalent

Answer 1

In this case $$\|f\|_{L^{1}}\leq \|f\|_{L^{\infty}} |B|$$ where $|B|$ denotes the volume of the compact set $B$.

Now, to show the norms are not equivalent we have to show that the inequality $$\|f\|_{L^{\infty}}\leq C \|f\|_{L^{1}} \qquad (1) $$ is not true for any constant $C$ independent of $f$.

Simply consider the functions $f_{n}(x)= \frac{1}{\sqrt{x+\frac{1}{n}}}$ on the compact set $B=[0,1]$. Then $\|f_n\|_{\infty}=\sqrt{n}$. But $\|f_n\|_{L^1} \sim \frac{1}{\sqrt{n}}$.

Notice that $\|f_n\|_{L^1}=2(\sqrt{n+1}-\sqrt{n})=2\frac{1}{\sqrt{n+1}+\sqrt{n}}<\frac{1}{\sqrt{n}}$. More imprtantly, we have shown that $$\|f_n\|_{L^{\infty}}/\|f_n\|_{L^{1}}>n.$$ Therefore (1) is not true with a constant $C$ independent of $f$.

Answer 2

Your idea $f_k(x) = 1-k x$ for $0\le x \le \frac{1}{k}$, and $0$ on $[\frac{1}{k}, 1]$ works OK for the set $B=[0,1]$. Indeed, we have $\|f_k\|_1 = \frac{1}{2k}$, and $\|f_k\|_{\infty} = 1$. Now, to consider the case $B= [0,1]^n$, you can take $F_k(x_1, \ldots, x_n) = f_k(x_1)$, and again $\|F_k\|_{1} = \frac{1}{2k}$, $\|F_k\|_{\infty} = 1$. Now you can do a similar thing for every cube.

Answer 3

Here's how to come up with a proof on your own by drawing pictures.

Note $\|f\|_1$ is the area under the graph of $|f|$, and $\|f\|_{\infty}$ is the max height of the graph of $|f|$.

If the two norms are equivalent, then there are constants $c,C>0$ such that $$c\|f\|_{1} \leq \|f\|_{\infty} \leq C \|f\|_{1}$$ for all $f$ in $C(B)$. You want to prove this does not happen.

Take $B = [0,1]$.

Draw the graph of a non-negative continuous function whose area underneath is 1 and whose max height is 10. Now graph of a non-negative continuous function whose area underneath is 1 and whose max height is 100. Finally, draw the graph of a non-negative continuous function whose area underneath is 1 and whose max height is 1,000,000.

Using what you learned from drawing graphs, show the following: For arbitrary $C > 0$, there exists a non-negative continuous function $f$ such that $\|f\|_{\infty} > C\|f\|_1$. That's the proof.