$||f|| = \int_0^1 |f(x)| dx$ and $||.||_{\infty}$ are not equivalent norms

Question

Show that the following function defines a norm in $C[0,1]$ which is not equivalent to the norm $||.||_{\infty}$:

$$||f|| = \int_0^1 |f(x)| dx$$

$$\int_0^1 ||f(x)|| = 0\implies f(x) = 0\mbox{ for all $x\in [0,1]$}$$

The triangle inequality follows from the triangle inequality from $|a+b|\le |a|+|b|$ but with function images in place of the absolute value number. Also $\int_0^1|\alpha f(x)| dx = \alpha \int_0^1|f(x)|\ dx$ so it is indeed a norm.

Now, $||f(x)||_{\infty} = \sup_{x\in [0,1]} \{f(x)\}$

In order to show that it's not equivlent to the norm $||.||$, I must show there are no constants $a,b>0$, such that

$$a||f||_{\infty} \le ||f||\le b||f||_{\infty}$$

Suppose that there are, so we have

$$a\sup_{x\in [0,1]}f(x) \le \int_0^1 |f(x)| dx\le b\sup_{x\in [0,1]}f(x)$$

I think

$$\int_0^1 |f(x)| dx\le \int_0^1 \sup_{x\in [0,1]} f(x) \ dx = \int_0^1 \ dx \sup_{x\in [0,1]} f(x) = b \sup_{x\in [0,1]} f(x)$$

I think I can see $a\sup_{x\in [0,1]}f(x) \le \int_0^1 |f(x)| dx$ happening. Why it cannot be true for some very small $a$?

Answer 1

Let $f_n(x)=1-nx$ for $0<x<\frac 1 n$ and $0$ for all other $x$. What is $\|f_n\|_{\infty}$ and $\|f_n\|_1$ ?.

Answer 2

Hint: think about a function that is "mostly" zero except for a very sharp spike.