Let $p \in (1, \infty)$. If $E$ and $F$ are Banach spaces, we define $E \oplus_p F$ to be the usual direct sum equipped with the norm $\|(e,f)\|=(\|e\|^p+\|f\|^p)^{1/p}$.
Now let $(X_j, \mu_j)$ be measure spaces for $j=1,2$.
Suppose that $T_k: L_p(\mu_1) \to L_p(\mu_2)$ are bounded linear maps for $k=1,2$.
Define $T_*: L_p(\mu_1) \to (L_p(\mu_2) \oplus_p L_p(\mu_2))$ by $$ T_*f=(T_1f, T_2f). $$ Define $T^*: (L_p(\mu_1) \oplus_p L_p(\mu_1)) \to L_p(\mu_2)$ by $$ T^*(f,g)=T_1f + T_2g. $$
Problem: Show that $T_*$ and $T^*$ are bounded linear maps and compute their norms.
Progress: I think I have successfully shown that $T_*$ and $T^*$ are bounded linear with $$ \| T_*\| \leq (\|T_1\|^p+\|T_2\|^p)^{1/p}, $$ and $$ \| T^*\| \leq (\|T_1\|^q+\|T_2\|^q)^{1/q}, $$ where $q$ is the Hölder conjugate of $p$, i.e. $\frac{1}{p}+\frac{1}{q}=1$.
Questions: I suspect that we get equality for the norms. Do we? I've tried to prove it but I can't. I feel that I am either missing something obvious or that it might not be true. If we do not get equality, is there any hope to compute the norms?
Hope someone can help me here. Thanks.
You do not get equality in general. Let me explain for $T_*$ only and leave the one for $T^*$ to you - it can be shown in much the same way. The proof for the norm inequality is \begin{align}\|T_*f\|&=\|(T_1f,T_2f)\|\\ &=\sqrt[p]{\|T_1f\|^p+\|T_2f\|^p}\\ &\le\sqrt[p]{\|T_1\|^p\|f\|^p+\|T_2\|^p\|f\|^p}\tag{1}\\ &=\sqrt[p]{\|T_1\|^p+\|T_2\|^p}\|f\|\end{align} There is only one inequality here, so to address whether the norm of $T_*$ can be achieved one needs to check whether a function $f$ can always be found which makes $(1)$ an equality (at least in the limit); that is $\|T_1f\|=\|T_1\|\|f\|$ and $\|T_2f\|=\|T_2\|\|f\|$. But linear operators need not have their maximum norm on the same function. This can be easily seen if one takes $T_i$ to be matrices.
So to construct a counterexample, take $X_1=X_2=\{1,2\}$, so $L_p(\mu)=\mathbb{R}^2_p$, and $T_1=\begin{pmatrix}1&0\\0&0\end{pmatrix}$, $T_2=\begin{pmatrix}0&0\\0&1\end{pmatrix}$, and $\mathbf{f}=\begin{pmatrix}x\\y\end{pmatrix}$, then $$\|T_*\|=\sup_{\mathbf{f}\ne0}\frac{\|T_*\mathbf{f}\|}{\|\mathbf{f}\|}=\sup\frac{\sqrt[p]{|x|^p+|y|^p}}{\sqrt[p]{|x|^p+|y|^p}}=1$$ $$\|T_1\|=\sup\frac{\|T_1\mathbf{f}\|}{\|\mathbf{f}\|}=\sup\frac{|x|}{\sqrt[p]{|x|^p+|y|^p}}=1=\|T_2\|$$ $$\implies \sqrt[p]{\|T_1\|^p+\|T_2\|^p}=\sqrt[p]{2}$$
Thus one cannot hope to calculate $\|T_*\|$ unless $T_1$, $T_2$ are specified.