# If $f\in L^1[0,1]\cap L^2[0,1]$, then $\|f\|_1 \le \|f\|_2$.

This problem has two parts:

(a) If $f\in L^1[0,1]\cap L^2[0,1]$, then $\|f\|_1 \le \|f\|_2$.

(b) Use (a) to deduce that $L^2[0,1]$ is a subset of $L^1[0,1]$.

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