How to prove that $L^{\infty}([0,1]) \subset L^q([0,1]) \subset L^p([0,1])$?

Question

We consider $1 \leq p < q < +\infty$.

How we can show that $L^{\infty}([0,1]) \subset L^q([0,1]) \subset L^p([0,1])$.

I really don't know how to prove these inclusions. I just know that $m([0,1])=1$ for Lebesgue measure, but it doesn't really help me.

Someone could help me ? Thank you in advance.

Answer 1

Hint:

Using Hölder inequality

$$\|f\|_p^p= \int_0^1 |f|^p .1 d\mu \leq \left( \int_0^1 |f|^{p \frac{q}{p}}d\mu \right)^\frac{p}{q} \mu([0,1])^{1- \frac{p}{q}}= \left( \int_0^1 |f|^q d\mu \right)^\frac{p}{q}=\|f\|_q^p$$

Answer 2

No doubt there is some really clever Hoelder inequality trick to do this, but I can't remember it. On the other hand, a little common sense will do it. Let $f \in L^q$ and let $A = \{x : |f(x)| \le 1\}$, and $A' = [0,1]\setminus A$.
$$ \int |f|^p = \int_A |f|^p + \int_{A'} |f|^p \le \mu(A) + \int_{A'} |f|^q \le 1 + ||f||_q^q, $$ where the first inequality uses that $|f(x)| \le 1$ for $x \in A$ and $|f(x)|^p \le |f(x)|^q$ for $x \in A'$, since $p < q$.