1
$\begingroup$

My question is:

Is there monotone property of $\|f\|_p$ when $p$ is increasing, where $\|f\|_p=(\int_a^b f(x)^pdx)^{1/p}$ is the classical $L^p$ norm and $f\in L^p$?

.

This proposition is right when $p\rightarrow\infty$ because $\|f\|_\infty=\max|f(x)|$. But I am interested in the case for which $p$ is finite.

My try:

Personally, the $\|f\|_p$ is increasing when $p$ is increasing. I use some common functions, such as $\sin(x)$, $\exp(x)$ etc., and they show that is valid. My idea is to prove the derivative of $\|f\|_p$ to $p$ is positive.

Let $g(p)=(\int_a^bf(x)^pdx)^{1/p}$. We have: $$ g'(p)=\dfrac{(\int_a^bf(x)^pdx)^{1/p-1}}{p}(\int_a^bf(x)^p(\ln f(x)^p-\ln\int_a^bf(t)^pdt)dx) $$ where $g'(p)$ is the derivative of $g(p)$ to $p$. Obviously, we only need to proof the 2nd item of $g'(p)$ is positive. To this end, let $h(x)=\ln(x)$ and $c=\int_a^bf(x)^pdx$, then from Taylor's expansion, we have: $$ h(x)=h(c)+h'(c)(x-c)+h''(\xi)(x-c)^2\leq h(c)+h'(c)(x-c)=h(c)+\dfrac{x-c}{c} $$ By setting $x=f(x)^p$ and integral from $a$ to $b$, we have: $$ \int_a^b\ln f(x)^pdx\leq (b-a)\ln\int_a^b f(t)^pdt+1-(b-a) $$ This is very close to $g'(p)$ by using mid point theorem. But I can not proceed. Does anyone can give me a ingenious proof or show me my proposition is wrong?

$\endgroup$
2
$\begingroup$

First, about your proof, you have no idea that the functions are differentiable. For example $1_{\mathbb{Q}\cap(0,1)}$ isn't differentiable in the traditional sense but it is $L^1(0,1) \cap L^2(0,1)$. Likewise, you implicitly specified the interval $(a,b)$ to be the set you're working on, but note that you cannot set $b=\infty$ since then $L^q(S) \subset L^p(S)$ when $p \le q$ and $S$ doesn't have arbitrarily large measure. (Embeddings).

Now, for a counterexample to your claim; Let $$ f(x) = 1/x $$ then $f \in L^1(1,3) \cap L^2(1,3)$ and $$ \int_1^3 \frac{1}{x} \, \mathrm{d} x = \ln(3) \\ \sqrt{\int_1^3 \frac{1}{x^2} \, \mathrm{d} x} = \sqrt{\frac{-1}{3} + \frac{1}{1}} = \sqrt{\frac{2}{3}} $$ but $$ \sqrt{\frac{2}{3}} < 1 < \ln(3) $$ so that $$ \|f\|_1 > \|f\|_2 $$ breaking your claim.

The key here seems to be that the exponent makes the integrand even smaller, more so than when one takes the square root afterwards.

$\endgroup$
  • $\begingroup$ $||f||_1>||f||_2$ is breaking his claim $\endgroup$ – Seyhmus Güngören May 24 '14 at 12:56
  • $\begingroup$ @SeyhmusGüngören that's a good point, I believe I messed up the inequality. $\endgroup$ – DanZimm May 24 '14 at 12:56
  • $\begingroup$ Thank you very much for your answer. But can you explain more about your explain about 'but note that you cannot set $b=\infty$ since...'? Do you mean that my method is invalid when an improper integral is considered? $\endgroup$ – Lion May 24 '14 at 13:13
  • $\begingroup$ @Lion correct, as is shown in my link, if you work on a set with arbitrarily large measure then the embedding does not exist. The real problem with your approach, in my opinion, however, is that you differentiate. $\endgroup$ – DanZimm May 24 '14 at 13:15
  • $\begingroup$ OK. Thank you for your answer. It is very helpful! $\endgroup$ – Lion May 24 '14 at 13:18
3
$\begingroup$

The inequality is true when when $L^p = L^p(S, \Sigma, \mu)$ is defined with respect to a probability measure $\mu$. For example, for $a < b$ and $p \leq q$,

$$ \begin{align} \frac{1}{a-b}\int_a^b{|f(x)|^p dx} &= \frac{1}{a-b}\int_a^b{(|f(x)|^q)^{p/q} dx}\\ &\leq \left(\frac{1}{a-b}\int_a^b{|f(x)|^q dx}\right)^{p/q} \end{align} $$ by Jensen's inequality and the concavity of the function $y \to y^{\alpha}$ for $\alpha = p/q \leq 1$. Raise both sides to the power $1/p$ and you get

$$ \left(\frac{1}{a-b}\int_a^b{|f(x)|^p dx}\right)^{1/p} \leq \left(\frac{1}{a-b}\int_a^b{|f(x)|^q dx}\right)^{1/q}. $$

So the point is that the inequality works when you normalize the measure the right way.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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