$T:L^2(0,1)\to L^2(0,1)$, is a compact operator. Show that $$ \forall \epsilon>0, \exists C_\epsilon>0; \forall f \in L^2(0,1), \|Tf\|_2 \leq \epsilon \|f\|_2+C_\epsilon\|f\|_1. $$
where the $\|f\|_2,\|f\|_1$is the $p-$norm.
We have the consequence that if a sequence $\{f_n\}_{n=1}^\infty$ in $L^2(0,1)$ satisfies $$ \|f_n\|_1 \to 0, \|f_n\|_2=1(n\geq1) $$ then $f_n \to 0$ weakly in $L^2(0,1)$.
Suppose this is false. Then there exists $\epsilon >0$ and a sequence $(f_n)$ such that $\|Tf_n\|_2 >\epsilon \|f_n\|_2 +n\|f_n||_1$. Let $g_n=\frac {f_n} {\|f_n\|_2}$. Then $\|g_n\|_2=1$ and $\|g_n\|_1 \leq \frac 1 n \frac {\|Tf_n\|_2} {\|f_n\|_2}$. Since $T$ is bounded it follows that $\|g_n\|_1 \to 0$. Hence $g_n \to 0$ weakly. But $T$ is compact, so $Tg_n \to 0$ in the norm. But $\epsilon <\|Tg_n\|_2$, a contradiction.