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This question is probably obvious, but I'm not seeing how to obtain it.

A simple function is said to be finitely simple if its support is of finite measure. Let $(X_1,\mu_1)$, $(X_2,\mu_2)$, and $(Y,\nu)$ be $\sigma$-finite measure spaces. Suppose I have a family $\{T_z : z \in S\}$ where $S = \{z \in \mathbf{C} : 0 \leq \mbox{Re}z \leq 1\}$, of bilinear operators defined on the space of finitely simple functions on $X_1 \times X_2$ and taking values in the space of measurable functions on $Y$ such that $$ \int_Y |T_z(\chi_{A_1},\chi_{A_2})|\chi_{B} d\nu < \infty $$ for all measurable subsets $A_1\subset X_1, A_2 \subset X_2, B\subset Y$ of finite measure. Suppose that I know there is an $\alpha \in (0,\pi)$ so that for all measurable subsets $A_1\subset X_1, A_2 \subset X_2, B\subset Y$ of finite measure $$ \log\left|\int_B T_z(\chi_{A_1},\chi_{A_2})d\nu \right| \leq C e^{\alpha |\mbox{Im}z|} $$ holds for some constant $C = C(A_1,A_2,B)$. Why does it then follow that for any two finitely simple functions $f_i$ on $X_i$ and finitely simple functions $g$ on $Y$ that $$ \log\left|\int_Y T_z(f_1,f_2) g \mbox{ }d\nu \right| \leq K e^{\alpha |\mbox{Im}z|} $$ for some constant $K = K(f_1,f_2,g)$? This seems like it should be really obvious by applying the linearity of the operator and of the integral, but logarithms are not subadditive so I'm not sure. Thanks in advance!

Note: For anyone wondering, I'm looking at admissible growth and multilinear complex interpolation results and ran across this statement.

Edit: Please see the solution below and let me know what you think. Thanks!

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Suppose $$ f_1 = \sum_{i = 1}^{m} \alpha_i \chi_{A_i}, \quad f_2 = \sum_{i = 1}^{m} \beta_i \chi_{B_i}, \quad g = \sum_{i = 1}^{m} \gamma_i \chi_{C_i}, $$ where we can assume each sum has the same number of terms by possibly taking some coefficients to be zero. Therefore, by the linearity of $T$ $$ \begin{align*} \left |\int_Y T(f_1,f_2)g\, dx \right| &= \left| \sum_{1 \leq i,j,k \leq m} \alpha_i \beta_j \gamma_k \int_{C_k} T(\chi_{A_i}, \chi_{B_j}) \, dx\right|\\ &\leq \sum_{1 \leq i,j,k \leq m} |\alpha_i \beta_j \gamma_k| \left|\int_{C_k} T(\chi_{A_i}, \chi_{B_j}) \, dx\right|\\ &\leq \sum_{1 \leq i,j,k \leq m} |\alpha_i \beta_j \gamma_k| e^{C(A_i,B_j,C_k) e^{\alpha \left|\mbox{Im z}\right|}} \end{align*} $$ which by setting $$ M = \max\{C(A_i,B_j,C_k) : 1\leq i,j,k \leq m\} \quad\mbox{ and } \quad N = \sum\limits_{1 \leq i,j,k \leq m} |\alpha_i \beta_j \gamma_k|, $$ we get $$ \left |\int_Y T(f_1,f_2)g\, dx \right| \leq Ne^{M e^{\alpha \left|\mbox{Im z}\right|}} . $$ Now take the logarithm of both sides to get $$ \log\left |\int_Y T(f_1,f_2)g\, dx \right| \leq e^{\alpha \left|\mbox{Im}z\right|} \big(e^{-\alpha \left|\mbox{Im}z\right|}|\log(N)| + M\big) \leq \big(|\log(N)| + M) e^{\alpha \left|\mbox{Im}z\right|} $$ and since $M$ and $N$ are constants depending on $f_1,f_2,g$, the desired result is obtained with $K = |\log(N)| + M$.

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