If $f \in L_4([0,1])$ then $f \in L_2([0,1])$ and $||f||_2 \leq ||f||_4$

Question

If $f \in L_4([0,1])$ then $f \in L_2([0,1])$ and $||f||_2 \leq ||f||_4$

I am not sure how to prove the first statement, we say that $f \in L_P$ if $\int |f|^p < \infty$. Then if $f \in L_4([0,1])$ then $\int |f|^4 < \infty$, but since $|f|^2< |f|^4$, then $\int |f|^2< \int |f|^4 < \infty$. Is this correct?

For the second part, so far I 've got $(\int |f|^2)^{\frac{1}{2}} \leq (\int |f|^4)^{\frac{1}{4}} $ iff $(\int |f|^2)^{2} \leq \int |f|^4 $, but from there I don't go anywhere. I thought using Cauchy-Schwarz would help. Can someone give me a hint?

Thank you in advance.

Answer 1

Write $|f|^2 = |f|^2 \cdot 1$ and apply the Cauchy-Schwarz inequality to get $\||f|^2\|_1 \le \||f|^2\|_2 \|1\|_2$. Simplify the inequality.

Note if $f \in L_4([0,1])$, then the inequality $\|f\|_2 \le \|f\|_4$ implies $\|f\|_2 < \infty$ and hence $f \in L_2([0,1])$.

Answer 2

It is certainly not true that $|f|^2 < |f|^4$ (think of the case where $|f|<1$).

Hint: Cauchy-Schwarz on $\int |f|^2 = \int g h$ where $g = |f|^2$ and $h = 1$.