Is it true that $\|f\|_p\le c\|f\|_r$ when $r<p$?

Question

Is it true that if $p,r\in [1,\infty]$, $r<p$ then $\exists c>0$ such that $\|f\|_p\le c\|f\|_r$ for all $f\in C_{[0,1]}$?

A friend of mine told me that it was false but I haven't been able to find a counterexample. I would really appreciate your help.

Answer 1

I give to you the case $r=1$ and $p=2$, and I leave to you to think of the general case.

To show that there does not exist a constant $c$ such that $$ \|f\|_2\leq c\|f\|_1,\quad\forall f\in C[0,1], $$ it suffices to show that given $A>0$ we can find a positive continuous function $f$ such that $$ \frac{\int_0^1 (f(x))^2\,dx}{\Bigl(\int_0^1 f(x)\,dx\Bigr)^2}>A. $$

Let $$ f_n(x)= \begin{cases} 2n^2x & 0\leq x<1/(2n)\\ 2n-2n^2 x & 1/(2n)\leq x<1/n\\ 0 & 1/n\leq x\leq 1 \end{cases} $$ Draw a figure! Then, $$ \int_0^1 f_n(x)\,dx=\frac{1}{2}, $$ and $$ \int_0^1 f_n(x)^2\,dx=\frac{n}{3}. $$ Hence, $$ \frac{\int_0^1 (f_n(x))^2\,dx}{\Bigl(\int_0^1 f_n(x)\,dx\Bigr)^2}=\frac{4n}{3}. $$ It is clear that this can be made larger than any given $A$ by choosing $n$ sufficiently large.

Answer 2

It should be reverse. By Holder's inequality, there is $c>0$ $$ (\|f\|_r)^r=\int_E |f|^r \:d\mu\leqslant \left(\int_E (|f|^r)^{p/r} \:d\mu\right)^{r/p}\left(\int_E 1^{p/(p-r)} \:d\mu\right)^{1-r/p}=(\|f\|_{p})^rM $$ Where $M=\mu(E)^{1-r/p}$. So let $c=M^{1/r}$. We have $$ \|f\|_r\leqslant c\|f\|_{p} $$ For a counter example, choose $p=3, r=1$ and $$ f_n(x)=\begin{cases}n^3x, \quad x\in[0,\frac1{n}] \\n^3(\dfrac2{n}-x), \quad x\in(\frac1{n},\frac2{n}] \\0,\quad x\in(\frac2{n},1] \end{cases} $$ Then $$ \|f_n\|_1=\int_0^1 f_n(x) \:dx=n\quad\text{and} \quad \|f_n\|_3=\left(\int_0^1 f_n^3(x) \:dx\right)^{1/3}=\frac{n^{5/3}}{\sqrt[3]{2}}=\frac{n^{2/3}}{\sqrt[3]{2}}\|f_n\|_1 $$ So for any $c>0$, there is $n$ such that $$\|f_n\|_3>c\|f_n\|_1$$

Answer 3

Consider the sequence of functions $(f_n)$ defined as follows: $f_n(x) = 0$ for $x\in[1,\frac{1}{2}-\frac{1}{n}]$ and $x\in[\frac{1}{2}+\frac{1}{n},1]$, and for $x\in(\frac{1}{2}-\frac{1}{n},\frac{1}{2}+\frac{1}{n})$ the graph of $f(x)$ traces the isoceles triangle with base along $(\frac{1}{2}-\frac{1}{n},\frac{1}{2}+\frac{1}{n})$ and of height $n$. In particular, $f_n(\frac{1}{2}) = n$. Then each $f_n$ is continuous (in fact, piecewise continuous and linear), and $\|f_n\|_1 = 1$ (the area of this isoceles triangle). But $\|f_n\|_\infty = n$. So there can be no single constant $c$ such that $$ \|f\|_\infty \leq c\|f\|_1 $$ for all $f\in C[0,1]$, because no finite $c$ can satisfy $$ n = \|f_n\|_\infty \leq c\|f_n\|_1 = c $$ for all $n$.