# Convergence in $L^{p_2}$ implies convergence in $L^{p_1}$

Let $\mu(\Omega)<\infty$ and $1\leq p_1\leq p_2\leq \infty$. Show that if $f_n \to f$ in $L^{p_2}$ then $f_n\to f$ in $L^{p_1}$.

I have a few ideas, based on my little experience of seeing other proofs:

• use Holder's inequality (somehow)
• considering that if $|f_n-f|>1$, then $|f_n-f|^{p_2}\geq |f_n-f|^{p_1}$

These two seem to be quite popular tricks for proving such inequalities, however, to use them concretely I have no idea..

I tried $\|f_n-f|_{p_1}^{p_1}=\int |f_n-f|^{p_1}d\mu\leq (\int |f_n-f|^{p_1p_2} d\mu)^\frac{1}{p_2}(\int 1^{q_2} d\mu)^\frac{1}{q_2}$ where $1/p_2+1/q_2=1$ by Holder's inequality, but that seems to be a dead end.

Thanks for any help.

• On the right hand side, you should try to get an integral $\int \lvert f_n - f\rvert^{p_2}\,d\mu$, because that is something you know a bit about. – Daniel Fischer Oct 20 '15 at 14:47
• Hint: $$\int |f_n - f|^{p_2} \,d\mu = \int (|f_n-f|^{p_1})^{p_2/p_1}\,d\mu$$ – Nate Eldredge Oct 20 '15 at 14:50

$$\| |f_n - f|^{p_1} \cdot 1 \|_1 \leq \| |f_n-f|^{p_1} \|_{p_2/p_1} \| 1 \|_{(p_2/p_1)'}$$
where $p'=\frac{p}{p-1}$ is the Holder conjugate. The point of this is that the first term on the right side is closely related to $\| f_n - f \|_{p_2}$, which is the thing you know something about, while the left side is closely related to $\| f_n - f \|_{p_1}$, which is the thing you want to control.