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?


2 Answers 2


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 $p \leq q$,

$$ \begin{align} \frac{1}{b-a}\int_a^b{|f(x)|^p dx} &= \frac{1}{b-a}\int_a^b{(|f(x)|^q)^{p/q} dx}\\ &\leq \left(\frac{1}{b-a}\int_a^b{|f(x)|^q dx}\right)^{p/q} \end{align} $$ 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

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

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

  • $\begingroup$ I know this is old, but did you want b-a?, lest the inequalities be reversed. $\endgroup$ Dec 22, 2019 at 18:13
  • $\begingroup$ @tangentbundle thanks, this was a typo $\endgroup$ Dec 22, 2019 at 21:17

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.

  • $\begingroup$ $||f||_1>||f||_2$ is breaking his claim $\endgroup$ May 24, 2014 at 12:56
  • $\begingroup$ @SeyhmusGüngören that's a good point, I believe I messed up the inequality. $\endgroup$
    – DanZimm
    May 24, 2014 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, 2014 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, 2014 at 13:15
  • $\begingroup$ OK. Thank you for your answer. It is very helpful! $\endgroup$
    – Lion
    May 24, 2014 at 13:18

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