If $f\not\in L^{\infty}$ then $\lim_{p\rightarrow\infty}||f||_p=\infty $.

Question

I want to prove that

If $f\not\in L^{\infty}$ then $\lim_{p\rightarrow\infty}||f||_p=\infty $.

I'd like any hint that approaches to solution.

I now that for every real number $r$ the set $\{x:|f(x)|>r\}$ has positive measure $f$ but I do not know what to do next.

Answer 1

If $f \not\in L^{\infty}$, then given any $M > 0$, we have $|f| > M$ on a set $E$ of positive measure. Therefore, $|f|^p > M^p$ on $E$, so $$\int |f|^p \geq \int_E |f|^p \geq M^p \mu(E)$$ If $\mu(E) = \infty$ then this shows that $\|f\|_p = \infty$ for all $p<\infty$, so the result certainly holds.

Otherwise, $$\|f\|_p = \left(\int |f|^p\right)^{1/p} \geq M \mu(E)^{1/p}$$ As $0 < \mu(E) < \infty$, we have $\lim_{p \to \infty}\mu(E)^{1/p} = 1$, so $$\liminf_{p \to \infty}\|f\|_p \geq M$$ As this holds for arbitrarily large $M$, we conclude that $\lim_{p \to \infty}\|f\|_p = \infty$.

Answer 2

Here's a hint: Use Chebyshev's Inequality to get that $$ \int\limits_X|f|^p\,dx\geq N^p~|\{x~|~|f(x)|>N\}| $$ true for every $N\in\mathbb N$ and $p>1$. From this observation it shouldn't be too difficult. Let me know if you would like the full details.