Assume $m(E) < \infty$. For $f \in L^\infty(E)$, show that $\lim_{n \to \infty}||f||_n= ||f||_\infty$
This problem comes from Royden and Fitzpatrick's Real analysis. I think the statement is false, but hopefully someone will correct me if I am mistaken. Let $E=[0,1/2]$, and consider $f(x)=x \in L^\infty(E)$. Since $f$ is continuous, $||f||_\infty = ||f|||_{max} = 1/2$. However,
$$||f||_n \le [m(E)]^{1/n} ||f||_\infty = (1/2)^{1/n}||f||_\infty,$$
which implies
$$\lim_{n \to \infty} ||f||_n \le ||f||_\infty \lim_{n \to \infty} (1/2)^n = 0,$$
and therefore $\lim_{n \to \infty} ||f||_n = 0 \neq 1/2$. What am I missing?
EDIT:
Made a stupid error: the "$n$" should be replaced with "$1/n$". I've been working on this problem for quite some time now; I could use a gentle prod in the right direction.
Denote $\left|E\right|=m(E)$. Thanks to Holder's inequality, $$ \left\|f\right\|_n^n=\int_E\left|f\right|^n{\rm d}\mu\le\left(\int_E\left|f\right|^{n\frac{n+1}{n}}{\rm d}\mu\right)^{\frac{n}{n+1}}\left(\int_E\left|1\right|^{n+1}{\rm d}\mu\right)^{\frac{1}{n+1}}=\left\|f\right\|_{n+1}^n\left|E\right|^{\frac{1}{n+1}}. $$ Therefore, $$ \left|E\right|^{-\frac{1}{n}}\left\|f\right\|_n\le\left|E\right|^{-\frac{1}{n+1}}\left\|f\right\|_{n+1}, $$ meaning that $\left|E\right|^{-\frac{1}{n}}\left\|f\right\|_n$ is monotonic (this indicates the existence of $\lim_{n\to\infty}\left\|f\right\|_n$). Thanks to this fact, $$ \left\|f\right\|_n\le\left\|\left\|f\right\|_{\infty}\right\|_n=\left\|f\right\|_{\infty}\left\|1\right\|_n=\left\|f\right\|_{\infty}\left|E\right|^{\frac{1}{n}}\iff\left|E\right|^{-\frac{1}{n}}\left\|f\right\|_n\le\left\|f\right\|_{\infty} $$ implies that $$ \lim_{n\to\infty}\left|E\right|^{-\frac{1}{n}}\left\|f\right\|_n\le\left\|f\right\|_{\infty}, $$ or, taking $\lim_{n\to\infty}\left|E\right|^{-\frac{1}{n}}=1$ into account, $$ \lim_{n\to\infty}\left\|f\right\|_n\le\left\|f\right\|_{\infty}. $$
On the other hand, $\forall\,\epsilon>0$ (where we require that $\epsilon<\left\|f\right\|_{\infty}$), $\exists\,A_{\epsilon}\subseteq E$, such that $$ \left|f\right|\ge\left\|f\right\|_{\infty}-\epsilon $$ holds on $A_{\epsilon}$. Therefore, we have $$ \left\|f\right\|_n\ge\left\|f\cdot 1_{A_{\epsilon}}\right\|_n\ge\left\|\left(\left\|f\right\|_{\infty}-\epsilon\right)1_{A_{\epsilon}}\right\|_n=\left(\left\|f\right\|_{\infty}-\epsilon\right)\left\|1_{A_{\epsilon}}\right\|_n=\left(\left\|f\right\|_{\infty}-\epsilon\right)\left|A_{\epsilon}\right|^{\frac{1}{n}}. $$ Consequently, $$ \left|A_{\epsilon}\right|^{-\frac{1}{n}}\left\|f\right\|_n\ge\left\|f\right\|_{\infty}-\epsilon, $$ which implies that $$ \lim_{n\to\infty}\left|A_{\epsilon}\right|^{-\frac{1}{n}}\left\|f\right\|_n\ge\left\|f\right\|_{\infty}-\epsilon. $$ Provided that $\lim_{n\to\infty}\left|A_{\epsilon}\right|^{-\frac{1}{n}}=1$, $$ \lim_{n\to\infty}\left\|f\right\|_n\ge\left\|f\right\|_{\infty}-\epsilon. $$ Due to the arbitrariness of $\epsilon$, this last inequality yields $$ \lim_{n\to\infty}\left\|f\right\|_n\ge\left\|f\right\|_{\infty}. $$
To sum up, we obtain $$ \lim_{n\to\infty}\left\|f\right\|_n=\left\|f\right\|_{\infty}. $$