let $1<p_1<p<p_2<\infty$. Let $Y$ be a Banach space and let $T$ be a bounded linear operator from $L^{p_1}$ to $Y$ and from $L^{p_2}$ to $Y$. Show that $T$ is then also a bounded linear operator from $L^p$ to Y.

I'm trying to use the lemma which states that for any $u\in L^p$ there exist $u_1\in L^{p_1}$ and $u_2\in L^{p_2}$ such that $u=u_1+u_2$.

So there exists $C_1$ such that $$||Tx||_y\leq C_1||x||_{p_1}$$ for any $x\in L^{p_1}$.

Also there exists $C_2$ such that $$||Tx||_y\leq C_2||x||_{p_2}$$ for any $x\in L^{p_2}$.

Hence if $s\in L^p$ then we can write $s=u_1+u_2$ such that $u_1\in L^{p_1}$ and $u_2\in L^{p_2}$. Then $$||Ts||_y\leq||Tu_1||_y+||Tu_2||_y\leq C_1||u_1||_{p_1}+C_2||u_2||_{p_2}\leq C[||u_1||_{p_1}+||u_2||_{p_2}],$$ for some constant $C$.

I don't know if it's possible to have inequality like $$||u_1||_{p_1}+||u_2||_{p_2}\leq C^* ||u_1+u_2||_p \ ??$$

  • $\begingroup$ As is, no. However, you should be able to construct a pair $(u_1,u_2)$ such that $\|u_1\|_{p_1} \leq C’_1\|x\|_p$ and same for $u_2$. Do not forget to show that the value of $T$ will not depend on the decomposition chosen! $\endgroup$ – Mindlack Jan 18 at 20:24

Assume $f\in L^p$ such that $\|f\|_p = 1$ is given. What we want to show is the existence of $C>0$ (which does not depend on $f$) such that $$ \|Tf\|_Y\le C. $$ Now, we can decompose $$ f=f1_{\{|f|\le 1\}}+f1_{\{|f|>1\}}=f_1+f_2. $$ Note that $\int |f_1|^{p_1}\le\int |f|^p= 1$ and $\int |f_2|^{p_2}\le\int |f|^p= 1$. Hence, $\|f_1\|_{p_1}\le$ and $\|f_2\|_{p_2}\le 1$ holds. Thus, we have $$ \|Tf\|_Y\le \|Tf_1\|_Y+\|Tf_2\|_Y\le C_1+C_2 $$ and the desired conclusion holds for $C=C_1+C_2$.

Note: In fact, Riesz-Thorin method (using complex method) provides a better bound $$ \|T\|_{p}\le \|T\|_{p_1}^\alpha\|T\|_{p_2}^{1-\alpha} $$ where $\frac{1}{p}=\frac{\alpha}{p_1}+\frac{1-\alpha}{p_2}.$

  • $\begingroup$ Oh, never mind. I misread the problem and was thinking of something more like Riesz-Thorin, where $T$ is bounded from $L^{p_1} \to L^{p_1}$ and from $L^{p_2} \to L^{p_2}$. Didn't realize that the codomain was the same space everywhere. $\endgroup$ – Nate Eldredge Jan 18 at 22:03
  • $\begingroup$ @NateEldredge Never mind! I think your concern is legitimate :) $\endgroup$ – Song Jan 18 at 22:04

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