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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)?

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You just proved the claim in part (a). –  Alex Becker Jan 6 '13 at 3:01
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up vote 3 down vote accepted

Let's review what you've shown. If $f\in L^2$, then $\|f\|_1\leq \|f\|_2$. This shows:

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

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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]) $.

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