I'm working on the following problem and got stuck. Any help would be really appreaciated. Let $a,b\in\mathbb{R}$ and $1\leq p \leq q < \infty$. Show that for any $f\in L^q([a,b])$ the following holds:

$$\frac{\lvert\lvert f\rvert\rvert_p}{(b-a)^{1/p}} \leq \frac{\lvert\lvert f\rvert\rvert_q}{(b-a)^{1/q}}$$

Thus $L^q([a,b])\subset L^p([a,b]).$

So I noticed that we could maybe use Hölder inequality with $g=\chi_{(a,b)}$, then $g$ is measurable and integrable, in particular:

$$g\in L^q,L^p$$ $$\lvert\lvert g\rvert\rvert_p = \left(\int\lvert g\rvert^p\right)^{1/p}= (b-a)^{1/p}$$ Then, if we suppose $f\in L^p$. By Hölder:

$$\lvert\lvert f \cdot g\rvert\rvert_1=\left(\int\lvert f\cdot g\rvert \right) \leq \lvert\lvert f\rvert\rvert_q\cdot \lvert\lvert g\rvert\rvert_p = \lvert\lvert f\rvert\rvert_q \cdot (b-a)^{1/p} $$

If we use Hölder again:

$$\frac{\lvert\lvert f \cdot g\rvert\rvert_1}{(b-a)^{1/p}} \leq \frac{\lvert\lvert f\rvert\rvert_p}{(b-a)^{1/p}}\cdot\lvert\lvert g\rvert\rvert_q = \frac{\lvert\lvert f\rvert\rvert_p}{(b-a)^{1/p}}\cdot (b-a)^{1/q} \leq \lvert\lvert f\rvert\rvert_q$$

Which would give us the inequality. But this would only be true if:

  1. $\frac{1}{q}+\frac{1}{p}=1$
  2. We don't know if $f\in L^p$

What am I missing or doing wrong? Thanks for the help.


1 Answer 1


Your last inequality seems obscure.
Anyway, perhaps you were hindered by the usual setting of Holder, so let me rephrase it for you : $$ \left( \int |h| \right)^m \left( \int |g| \right)^n \ge \left( \int |h|^{\frac{m}{m+n}}.|g|^{ \frac{n}{m+n}} \right)^{m+n}$$

for all measurable functions $f,g$ and positive real number $m,n$.
Your desired equality is the special case of Holder with : $$ h:= |f|^q ; g:= \mathbb{1}_{[a,b]} ; m=1$$ and $n>0$ satisies $ p =q\frac{m}{m+n}$ (note that $p<q$)

  • $\begingroup$ So with this setting it is okay to use $m=1$? $\endgroup$ Commented Nov 12, 2020 at 21:33
  • $\begingroup$ Yeah, because when you taking roots to the power of $(m+n)$ of both sides, you'll get the usual Holder's inequality. $\endgroup$ Commented Nov 12, 2020 at 21:35

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