# $\not\exists$ number $c\geq 0$ for which $||f||_{max}\leq c||f||_1$. But $\exists$ $c\geq 0$ for which $||f||_1\leq c||f||_{max}$

I'm reading through Real Analysis by Royden & Fitzpatrick and I always try to deal with the first few problems of each section. For the first section on $L^p$ spaces there were mostly routine exercises on proving norms, but this subquestion has me a bit confused:

Show that

a) $\not\exists$ number $c\geq 0$ for which $||f||_{max}\leq c||f||_1$ $\forall f \in C[a,b]$.

b) But $\exists$ $c\geq 0$ for which $||f||_1\leq c||f||_{max}$ $\forall f \in C[a,b]$

Where:

$||f||_{max}:= max_{x\in[a,b]}|f(x)|$

$||f||_1:=\int_a^b|f|$

EDIT: Does the following counterexample work for part a)?

Let $f(x)=x^n$ on $[0,1]$, $max|f(x)| > c \int_a^b|f(x)|$ leads to $1>|\frac{c}{n+1}|$ and for any c, however large, we can take n=c and that will keep our inequality true.

Hints:

• For nonexistence of $c$ you could try to construct non-negative continuous functions, which have a small integral, i.e., $\|f\|_1$ is small, but with large $\|f\|_{max}$.
• The second inequality follows with $c = b - a$.
• Thanks, the second one is silly I should've thought it through. Does this work for the first one: Thanks for your answer. So would something like this work?: Let $f(x)=x^n$ on$[0,1]$, $max|f(x)| > c \int_a^b|f(x)|$ leads to $1>|\frac{c}{n+1}|$ and for any c, however large, we can take n=c and that will keep our inequality true. – Mike Nov 30 '15 at 13:49
• Yes, that is fine. And it can be adopted to arbitrary intervals $[a,b]$. – gerw Nov 30 '15 at 14:03

The max norm is not the (absolute value of) the function at one specific point, but at a point that depends on the function itself. In a more general setting where functions are not necessarily continuous, it is called the supremum norm.

The integral of $|f|$ cannot be higher than the length of the integration interval multiplied by the maximum of $|f|$ because of the definition of the Riemann integral (the constant function $\max(|f|)$ defines an upper sum in the sense of the Riemann integral).

On the other hand, one can construct positive continuous functions on the interval that have very small surface areas underneath their graphs, yet their maximum is defined by a very sharp spike.

• Thanks for your answer. So would something like this work?: Let $f(x)=x^n$ on $[0,1]$, then $max|f(x)| > c \int_a^b|f(x)|$ leads to $1>|\frac{c}{n+1}|$ and for any c, however large, we can take n=c and that will keep our inequality true. – Mike Nov 30 '15 at 13:48
• That is a good example. Even on a smaller interval, even though the max would converge to 0, it would still do so by a factor $(n+1)$ more slowly than the integral, with the result that the ratio between the max and the integral becomes arbitrarily large. – Justpassingby Nov 30 '15 at 14:02