2010 February « Thoughts on my Mind

Correct “Infinity Norm” February 6, 2010

Posted by putnam120 in Uncategorized.
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A while back you might have seen that I wrote a post about the “infinity norm“. Well in that post I assumed that the measure of the entire space was finite, which was a good start but it doesn’t quite get the job done. So here I will present all (well almost all) of the necessary material to accomplish my desired goal.

To start things off, I will assume that you are familiar with Hölder’s Inequality, if not the follow the link and at least read everything before the “contents” section. So this is a common version of the inequality, however there is another inequality that can be derived from this (which one of my teachers call the Generalized Hölder’s Inequality). Its statement is as follows:

$\underline{\text{Generalized Holder:}}$

Suppose that $0<p<r<s\le\infty$ and that $f\in L^p$ as well as $f\in L^s$ . Then $f\in L^r$ , and if we choose $\lambda\in (0,1)$ such that $r^{-1}=\lambda p^{-1}+(1-\lambda)s^{-1}$ , we have that

$(1)\ \ \ \ \displaystyle \|f\|_r\le\|f\|_p^\lambda\|f\|_s^{1-\lambda}$

proof: First, if $s=\infty$ then we have that $|f|^r\le\|f\|_\infty^{r-p}|f|^p$ and $\lambda=\frac{p}{r}$ . Now if we integrate and take r-th roots we have

$\|f\|_r\le\|f\|_p^\lambda\|f\|_\infty^{1-\lambda}$

Seeing how we got that $\|f\|_p^\lambda$ takes a little thinking, but it shouldn’t be too difficult to convince yourself that this is correct.

Now assume that $s<\infty$ . Noticing that $r\lambda p^{-1}+r(1-\lambda)s^{-1}=1$ , we can use Hölder’s to obtain

$\displaystyle\int |f|^r=\int |f|^{r\lambda}|f|^{r(1-\lambda)}\le\||f|^{r\lambda}\|_{p/r\lambda}\||f|^{r(1-\lambda)}\|_{s/r(1-\lambda)}$ . This can be rewritten as ${\left(\int |f|^p\right)^{r\lambda/p}\left(\int |f|^s\right)^{r(1-\lambda)/s}=\|f\|_p^{r\lambda}\|f\|_s^{r(1-\lambda)}}$ . Then taking r-th roots gives us $(1)$ . ♦♦♦

Of all the things we will need to accomplish out goal, the above theorem was the most difficult to prove. Next we will prove what has been labeled “The world’s second most obvious inequality”. Don’t worry, this is almost as obvious as the “The world’s most obvious inequality” $(|\int f|\le\int |f|)$ .

$\underline{\text{Chebyshev's Inequality:}}$

Let $E_\alpha=\{x:|f(x)|>\alpha\}$ , then for $0<p<\infty$ and $f\in L^p$ , we have

$(2)\ \ \ \displaystyle\mu(E_\alpha)\le\left(\frac{\|f\|_p}{\alpha}\right)^p$

proof: It is almost insulting to write this out since it more or less proves itself. You just do the obvious thing and everything works itself out. However, for the sake of completeness I will provide the proof.

$\|f\|_p^p=\int |f|^p\ge\int_{E_\alpha} |f|^p\ge\int_{E_\alpha} \alpha^p=\alpha^p\mu(E_\alpha)$ . ♦♦♦

Now we finally have all the necessary tools to accomplish our goal.

$\underline{\text{Theorem:}}$ If $f\in L^p\cap L^\infty$ for some $0< p<\infty$ then $f\in L^q$ for all $q>p$ . Additionally,

$(3)\ \ \displaystyle\lim_{p\to\infty}\|f\|_p=\|f\|_\infty$

proof: By using $(1)$ we will show that $f\in L^q$ . Since $p<q<\infty$ , we can choose $\lambda\in (0,1)$ such that

$(4) \ \ \ q^{-1}=\lambda p^{-1}.$

Thus it follows that $\|f\|_q\le\|f\|_p^\lambda\|f\|_\infty^{1-\lambda}<\infty$ . So the first half of the theorem is proved.

Now since $p$ in $(4)$ is fixed, if $q\to\infty$ we must have that $\lambda\to 0$ . Hence

$\displaystyle\lim_{q\to\infty}\|f\|_q\le\lim_{\lambda\to 0}\|f\|_p^\lambda\|f\|_\infty^{1-\lambda}=\|f\|_\infty$

For the reverse direction we will make use of Chebyshev’s Inequality $(2).$ Before doing so however, if $\|f\|_\infty=0$ then we are done, so suppose that $\|f\|_\infty>0$ . Now choose $\epsilon>0$ such that $\|f\|_\infty-\epsilon>0$ . Let $\alpha=\|f\|_\infty-\epsilon$ , so $\alpha\mu(E_\alpha)^{1/p}\le\|f\|_p$ . Hence