# Inequality between $L^2$- and $L^1$-norms for functions

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.

## 2 Answers

It is not always true. For example, if $\mu$ is the Lebesgue measure on $[0,1]$ and $f(x) = x$, then $\|f\|_2 = \frac{1}{\sqrt{3}} > \frac{1}{2} = \|f\|_1$.

Generally, if $(X,\Sigma,\mu)$ is a finite measure space, and $f$ is a measurable function, then $\|f\|_1 \le \mu(X)^{1/2} \|f\|_2$. The estimate is obtained by applying the Cauchy-Schwarz inequality.

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 $$\| 1_E\|_1 = \mu(E)<\mu(E)^{1/2}= \|1_E\|_2.$$