TITLE: Inequality between $L^2$- and $L^1$-norms for functions
QUESTION [7 upvotes]: In the vector space $\mathbb{R}^n$, we have the inequality
$$
||x||_2 \leq ||x||_1
$$
where $x$ is a vector. 
I am wondering that we have similar inequality for function's norm.
The $L^1$-norm of function $f$ is given by
$$
||f||_1=\int |f| d\mu, 
$$
the $L^2$-norm is 
$$
||f||_2=\left( \int |f|^2 d\mu \right)^{1/2}
$$
then, is this right?
$$
||f||_2\leq ||f||_1
$$
If so, how to prove it? And tell me what I should study. e.g. real analysis or Lebesgue integral.

REPLY [3 votes]: Such an inequality can never hold for all positive functions and all measures. Consider scaling the measure by a positive constant $a>0$ (that is, let $\mu\mapsto a\cdot \mu$). Then the inequality turns into
$$
\|f\|_2\leq a^{1/2} \|f\|_1.
$$
Since the right hand side tends to zero as $a$ tends to zero, the inequality can only hold for all scaled measures if $f=0$ almost everywhere.
In fact, the inequality is false in any space which has a set $E$ with $0<\mu(E)<1$, since
$$
\| 1_E\|_1 = \mu(E)<\mu(E)^{1/2}= \|1_E\|_2.
$$