# $\|\cdot\|_1 \leq \|\cdot\|_2 \leq \|\cdot\|_{\infty}$ for functions in $C([0,1])$?

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

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

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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
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

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 $$\Vert f\Vert_2= \left(\int\limits_0^1|f(x)|^2dx\right)^{1/2}\leq \left(\int\limits_0^1\Vert f\Vert_\infty^2 dx\right)^{1/2}= \Vert f\Vert_\infty\left(\int\limits_0^1 1 dx\right)^{1/2}=\Vert f\Vert_\infty$$

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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 $p$.

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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 $C_1=C_2=1$.

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