Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Why does the following hold for continuous functions on $[0,1]$?

$\|\cdot\|_1 \leq \|\cdot\|_2 \leq \|\cdot\|_{\infty}$

share|improve this question
    
It isn't. E.g. $\| (1,1) \|_1=2$, $\| (1,1) \|_2=\sqrt{2}$, $\| (1,1) \|_{\infty}=1$. –  Chris Eagle May 11 '12 at 14:30
1  
Which spaces are you considering? –  Norbert May 11 '12 at 14:30
    
It's the other way, at least on $\mathbb K^d$. –  martini May 11 '12 at 14:31
    
Consider providing more context to this question. It will help get better answers. –  Pedro Tamaroff May 11 '12 at 14:31
    
Whats $\mathbb{K}^d $? –  rk101 May 11 '12 at 14:33
add comment

3 Answers

up vote 2 down vote accepted

From Holder's inequality (with $p=q=2$) one may deduce that for all $f\in C([0,1])$ $$ \Vert f\Vert_1= \int\limits_0^1|f(x)|dx= \int\limits_0^1|f(x)|\cdot |1| dx\leq \left(\int\limits_0^1|f(x)|^2dx\right)^{1/2}\left(\int\limits_0^1 |1|^2dx\right)^{1/2}= \Vert f\Vert_2 $$ Moreover $$ \Vert f\Vert_2= \left(\int\limits_0^1|f(x)|^2dx\right)^{1/2}\leq \left(\int\limits_0^1\Vert f\Vert_\infty^2 dx\right)^{1/2}= \Vert f\Vert_\infty\left(\int\limits_0^1 1 dx\right)^{1/2}=\Vert f\Vert_\infty $$

share|improve this answer
add comment

Since $x^{q/p}$ is convex for $q/p\ge1$, Jensen's Inequality yields $$ \|f\|_p^q=\left(\int_I|f(x)|^p\mathrm{d}x\right)^{q/p}\le\int_I|f(x)|^{pq/p}\mathrm{d}x=\|f\|_q^q $$ where $|I|=1$. Therefore, $\|f\|_p\le\|f\|_q$ when $p\le q$.

So on a space of measure $1$, $\|f\|_p$ is monotonically increasing in $p$.

share|improve this answer
add comment

I guess that the equation should be read as $$\|\cdot \|_1 \lesssim \|\cdot \|_2 \lesssim \|\cdot \|_\infty,$$ i.e. $$\|\cdot \|_1 \leq C_1 \|\cdot \|_2 \leq C_2 \|\cdot \|_\infty$$ for suitable constants $C_1>0$, $C_2>0$. In other words, $L^1 \supset L^2 \supset L^\infty$ with continuous immersions. This is true if you integrate on domains of finite measure, as follows from the Hölder inequality.

In the particular case of your question, it can be shown that you can take $C_1=C_2=1$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.