# Application of Hölder to prove $L^q \subset L^p$ for $1\leq p\leq q<\infty$ with Lebesgue measure

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.

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