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Let $f$ be a function of $L^p([0,2]) \>\> \forall p \in [1, \infty )$ and suppose $||f||_p \leq 1$. Show that $f$ belongs to $L^{\infty}([0,2])$ and $||f||_{\infty} \leq 1$.

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Suppose $|f| > 1$ on a subset of $[0,2]$ with positive measure, then since $\{x, |f(x)| > 1 \}= \cup_n \{x, |f(x)| > 1 +\frac{1}{n}\}$, there exists $n_0$, such that $A =\{x, |f(x)| > 1 +\frac{1}{n_0}\}$ is of positive measure, i.e. $\mu(A) >0$

Then we can see that \begin{align} ||f||_p \geq (\int_A |f(x)|^pdx)^{\frac{1}{p}} > (\int_A |1+\frac{1}{n_0}|^pdx)^{\frac{1}{p}} = (1+\frac{1}{n_0})\mu(A)^\frac{1}{p} \end{align}

Thus $\lim_{p\to \infty}||f||_p >\lim_{p\to \infty}(1+\frac{1}{n_0})\mu(A)^\frac{1}{p} = (1+\frac{1}{n_0})$. There exists some $N$ such that for all $p>N$, $||f||_p > 1+\frac{1}{2n_0}$. Thus contradiction with given conditions.

PS: if we know a priori $||f||_{\infty} < \infty$, we have $\lim_{p\to \infty}||f||_p = ||f||_{\infty}$. We have just proven this is also true when $||f||_{\infty} = \infty$

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Thank you very much. It is really clear. – user73793 May 25 '14 at 7:34

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