Exercise:
Assume $E$ has finite measure and $1\le p_{1}<p_{2}\le\infty$. Show that if {$f_{n}$}$\space\rightarrow f$ in $L^{p_{2}}(E)$ then {$f_{n}$}$\space\rightarrow f$ in $L^{p_{1}}(E)$
Proof:
Take $p_{1}<p_{2}$ then $||f||_{p_{1}}\le||f||_{p_{2}}$
$0 \le||f_n-f||_{p_{1}}\le||f_n-f||_{p_{2}} \rightarrow 0$
$\therefore ${$f_{n}$}$\space\rightarrow f$ in $L^{p_{1}}(E)$
Is this correct?
As pointed out by Daniel Fischer, the inequality $\lVert f\rVert_{p_1}\leqslant \lVert f\rVert_{p_2}$ doest not necessarily hold.
However, notice that for a fixed $\varepsilon$ \begin{align*}\int_E |f_n-f|^{p_1}dx&=\int_{\{|f_n-f|<\varepsilon\}}|f_n-f|^{p_1}dx+ \int_{\{|f_n-f|\geqslant \varepsilon\}}|f_n-f|^{p_1}dx\\ &\leqslant \varepsilon^{p_1}\mu(E)+\varepsilon^{p_1}\int_{\{|f_n-f|\geqslant \varepsilon\}}\left(\frac{|f_n-f|}{\varepsilon}\right)^{p_1}dx\\ &\leqslant \varepsilon^{p_1}\mu(E)+\varepsilon^{p_1}\int_{\{|f_n-f|\geqslant \varepsilon\}}\left(\frac{|f_n-f|}{\varepsilon}\right)^{p_2}dx\\ &\leqslant \varepsilon^{p_1}\mu(E)+\varepsilon^{p_1-p_2}\int |f_n-f|^{p_2}dx. \end{align*} We infer that $$\limsup_{n\to +\infty}\int_E |f_n-f|^{p_1}dx\leqslant \varepsilon^{p_1}\mu(E),$$ and we conclude, as $\varepsilon$ was arbitrary.