$L^1$ and $L^2$ norm inequaliy

Question

Consider real valued function $f$ defined on $[0, T]$.

L1 norm and L2 norm of function $f$ are given by

$$ \|f\|_1=\int_0^T |f(t)| \, dt $$ and $$ \|f\|_2=\sqrt{\int_0^T |f(t)|^2 \, dt } $$

Then we have the following inequality

$$\|f\|_2 \le \mu(X)^{-1/2} \|f\|_1.$$

at this inequality, what is $\mu(X)$??

$\mu(X)=T$??

Answer 1

If $X = [0,T]$ and $\mu$ is the Lebesgue measure on $X$, then by Cauchy-Schwarz $$\|f\|_1 \le \mu(X)^{1/2} \|f\|_2 = \sqrt{T}\|f\|_2$$ (Note: Your inequality states $\|f\|_1 \le \mu(X)^{-1/2}\|f\|_2$, which is not always true. For example, if $f(t) = 1$ and $T = 2$, then $\|f\|_1 = 2$ and $\mu(X)^{-1/2}\|f\|_2 = 1 < 2 = \|f\|_1$.)