# Bounded Linear Operators on Banach Spaces

let $$1. Let $$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 $$||u_1||_{p_1}+||u_2||_{p_2}\leq C^* ||u_1+u_2||_p \ ??$$

• 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! – 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 $$\frac{1}{p}=\frac{\alpha}{p_1}+\frac{1-\alpha}{p_2}.$$
• 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. – Nate Eldredge Jan 18 at 22:03