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

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

  • 0
    Whats $\mathbb{K}^d $?2012-05-11

3 Answers 3

2

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 $

2

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 $p$.

1

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 $C_1=C_2=1$.