Comparison of $L^2$ and $L^1$ norms for functions

Solution 1:

Yes, the inequality is given by Cauchy-Schwarz: $$ \|f\|_{L^1} = \int_a^b |f(x)|\, dx \le \left(\int_a^b |f(x)|^2\, dx\right)^{1/2}\left(\int_a^b \, dx\right)^{1/2} = \sqrt{b - a}\|f\|_{L^2}. $$

Solution 2:

Hint: Try using the Hölder inequality for integrals on the function $f\cdot 1$.

In general this will not be true. It is of great importance that your measure space (here: $[a,b]$) has finite measure. Then the $L^p$ spaces fit nicely into each other, i.e. $L^q\subseteq L^p$ whenever $p\le q$ (cf. in the section "embeddings" of https://en.wikipedia.org/wiki/Lp_space#Lp_spaces_and_Lebesgue_integrals for a general solution).