Limit of $L^p$ Norms and Essential Supremum 
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.
 A: 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
$$
\lim_{n\to\infty}\left\|f\right\|_n=\left\|f\right\|_{\infty}.
$$
