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 $L^1[0,1]$.
Without using part (a), let $f$$\in$$L^2[0,1]$. Since the constant function $1$ $\in$ $L^2[0,1]$, by Holder inequality, we can conclude that $\|f\|_1$$\le$$\|f\|_2\|1\|_2$$=\|f\|_2$.
But how can you deduce part (b) from part (a)? Also, how do you prove the claim in part (a)?
 A: Let's review what you've shown.  If $f\in L^2$, then $\|f\|_1\leq \|f\|_2$.  This shows:


*

*$L^2\subseteq L^1$, and therefore $L^1\cap L^2=L^2$.

*For all $f\in L^1\cap L^2 = L^2$, $\|f\|_1\leq \|f_2\|$.


I.e., you basically had proved (a) as well as (b) with the same step.
A: For Part (b), I cannot see immediately how it is implied by (a), but this is how you may proceed directly (just a suggestion, of course). Let $ f \in {L^{2}}([0,1]) $. Define
$$
A := \{ x \in [0,1] ~|~ |f(x)| \leq 1 \} \quad \text{and} \quad B := \{ x \in [0,1] ~|~ |f(x)| > 1 \}.
$$
For each $ x \in B $, we have $ |f(x)| < |f(x)|^{2} $, so
$$
\int_{B} |f|^{2} ~d{\mu} < \infty \Longrightarrow \int_{B} |f| ~d{\mu} < \infty.
$$
Next, notice that $ \displaystyle \int_{A} |f| ~d{\mu} < \infty $ because $ |f| $ is bounded by the value $ 1 $ on $ A $, which has finite measure. As $ \{ A,B \} $ partitions the interval $ [0,1] $, we see that
$$
\int_{[0,1]} |f| ~d{\mu} = \int_{A} |f| ~d{\mu} + \int_{B} |f| ~d{\mu} < \infty.
$$
Therefore, $ f \in {L^{1}}([0,1]) $.
