If $f \in L^{p_1}(E) $ is bounded then $f\in L^{p_2}$ for any $p_2>p_1$ 
Show that if $f$ is a bounded function on $E$ that belongs to $L^{p_1}(E)$ then it belongs to $L^{p_2}(E)$ for any $p_2>p_1$

How can I insert argument about the boundedness of $f$? I can prove $L^{p_2} \subset L^{p_1}$ but I am stuck here. Please help.
 A: Let $f \in L^{p_1}(E)$ bounded. Then there exists $M\geq 0$ such that $\sup_{x \in E} |f(x)| \leq M$ and $\|f\|_{p_1}^{p_1}=\int_E |f(x)|^{p_1}\,dx < \infty$.

Case 1: $p_2<\infty$
For $p_2 > p_1$, we have
\begin{align}
\|f\|_{p_2}^{p_2}& =\int_E |f(x)|^{p_2}\,dx \\
&= \int_E |f(x)|^{p_2-p_1}|f(x)|^{p_1}\,dx \\
& \leq \left(\sup_{x \in E} |f(x)|^{p_2-p_1}\right)\int_E |f(x)|^{p_1}\,dx \\
& \leq \left(\sup_{x \in E} |f(x)|\right)^{p_2-p_1}\int_E |f(x)|^{p_1}\,dx \\
& \leq M^{p_2-p_1} \|f\|_{p_1}^{p_1} < \infty
\end{align}
so that $\|f\|_{p_2}^{p_2} < \infty$ and hence $f \in L^{p_2}(E)$.
Note that we have used the fact that $p_2-p_1>0$ to claim $\sup_{x \in E} |f(x)|^{p_2-p_1}\leq \left(\sup_{x \in E} |f(x)|\right)^{p_2-p_1}$.

Case 2: $p_2=\infty$
Since $f$ is bounded, $\|f\|_{\infty}=\operatorname{ess sup}_{x \in E} |f(x)| \leq \sup_{x \in E} |f(x)| \leq M< \infty$ by hypothesis, so $f \in L^{\infty}(E)$.
A: Hint: If $|f|\le 1,$ then $|f|^{p_2} \le |f|^{p_1}.$

Further on the hint:
We have $f$ bounded and in $L^{p_1}.$ Let $g=f/\|f\|_\infty.$ Then $g\in  L^{p_1}.$ But since $|g|\le 1$ and $p_2 > p_1,$ 
$$|g|^{p_2} \le  |g|^{p_1}.$$
Thus $g \in L^{p_2},$ which implies $g\|f\|_\infty = f \in L^{p_2}.$
