Proof: All functions in $L^2[0,1]$ are in $L^1[0,1]$

Question

I would like to know if my demonstration of all functions in $L^2[0,1]$ are in $L^1[0,1]$?

• $\forall f\in L^2[0,1] $ we can split $f$ in two differents spaces: $A=\left \{0\leq x\leq 1:|f(x)|>1) \right \}$ and $\textrm{no}A=\left \{0\leq x\leq 1:0\leq |f(x)|\leq 1 \right \}$ (Of course $A \cup \textrm{no}A=[0,1]$.)

• We know too that $y^2>|y|$ iff $|y|>1$. So:

  1. $ \int_{\textrm{no}A}|f(x)|\mathrm{d}x \leq \int_{0}^{1}1\mathrm{d}x=1<\infty $
  2. $\infty>\int_{0}^{1}f^2(x)\mathrm{d}x\ge\int_{A}f^2(x)\mathrm{d}x> \int_{A}^{}|f(x)|\mathrm{d}x$

• So by (1)+(2): $ \int_{0}^{1}|f(x)|\mathrm{d}x = \int_{\textrm{no}A}|f(x)|\mathrm{d}x+\int_{A}|f(x)|\mathrm{d}x < \infty$ so by definition $f$ is in $L^1[0,1]$ too.

Is it correct?

Thank you.

Answer 1

You are overthinking albeit in a much elegant way.

Use the Cauchy-Schwarz inequality. You have that $\mathbf{1}_{[0,1]}\in L^{p}[0,1]\,, \forall p\geq 0$ . Let $f\in L^{2}[0,1]$

then $\int_{[0,1]}|f\cdot \mathbf{1}_{[0,1]}|\,d\lambda\leq \sqrt{\int_{[0,1]}|f|^{2}\,d\lambda}\cdot\sqrt{1} <\infty$ as $f\in L^{2}[0,1]$.