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:
- $\frac{1}{q}+\frac{1}{p}=1$
- We don't know if $f\in L^p$
What am I missing or doing wrong? Thanks for the help.
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 $p<q$)