if $\{f_n\}\to f$ in $L^{p_2}(E)$ then $\{f_n\}\to f$ in $L^{p_1}(E)$. Assume that $E$ has finite measure and $1 \le p_1 < p_2 \le \infty$. Show that if $\{f_n\}\to f$ in $L^{p_2}(E)$ then $\{f_n\}\to f$ in $L^{p_1}(E)$.
my attempt:  let $p=\frac{p_2}{p_1}$ and $1=\frac{1}{p}+\frac{1}{q}$,  and pick $n$ such that $\|f_n-f\|_{p_2}\le \epsilon , \forall n\ge N$
\begin{align}
\|f_n-f\|_{p_1}^{p_1} 
& = \int_E |f_n-f|^{p_2.\frac{p_1}{p_2}}d\mu = \int_\mathbb{R} |f_n-f|^{p_1.\frac{p_2}{p_1}} \chi_Ed\mu\\
& \le \|f_n-f\|_{p_2}^{p_1} . [\mu(E)]^{\frac{1}{q}} ~~~~~~\text{Holder}\\
\end{align}
so;
\begin{align}
\|f_n-f\|_{p_1} 
& = \|f_n-f\|_{p_2} . [\mu(E)]^{1-\frac{p_1}{p_2}} \\
& \le \epsilon \to 0\\
\end{align}

Good. But you should also take care of the case that $p_{2}=\infty$, but this is easy:
\begin{align*}
\|f_{n}-f\|_{L^{p_{1}}}^{p_{1}}=\int_{E}|f_{n}-f|^{p_{1}}\leq\|f_{n}-f\|_{L^{\infty}}^{p_{1}}\int_{E}=\|f_{n}-f\|_{L^{\infty}}^{p_{1}}\mu(E).
\end{align*}
