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Suppose $X$ and $Y$ are finite sets and $K:X\times Y\to \mathbb R$ is some function. We get an integral transform from the space of real functions on $X$ to real functions on $Y$ given by $$\Phi_Kf(y)=\sum_{x\in X}K(x,y)f(x).$$

We also have the adjoint which is going the other way: $$\Phi_K^*g(x)=\sum_{y\in Y}K(x,y)g(y).$$

The duality principle, which is often useful in number theory, says that these two transforms have the same norm when viewed as maps between $L^2(X)$ and $L^2(Y)$. Specifically, we have $$\sum_{y\in Y}(\Phi_Kf(y))^2\leq C\sum_{x\in X}f(x)^2$$ for any $f$ if and only if $$\sum_{x\in X}(\Phi_K^*g(x))^2\leq C\sum_{y\in Y}g(y)^2$$ for any $g$. Now suppose that for some $f:X\to\mathbb R$ we have a lower bound for $\sum_{y\in Y}(\Phi_Kf(y))^2$. The duality principle then tells us that we can deduce a corresponding lower bound for $\sum_{x\in X}(\Phi_K^*g(x))^2$ for some function $g:Y\to \mathbb R$. My question is then this: can we have any more information about the function $g$, perhaps with further assumptions on $K$? For instance, we might want to avoid $g$ being highly oscillatory, or too concentrated, etc.

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    $\begingroup$ I have not dealt with this exact setting, but we use a similar set up quite often in several complex variables. In that setting, it is often profitable to look at duality with respect to weighted $L^2$ norms, and tailor a family of weights to the problem at hand. Your question is so general that I do not feel I can say more than this. $\endgroup$ Commented Nov 19, 2014 at 22:13
  • $\begingroup$ In other words, an $n\times m$ matrix $K$ and its transpose have the same operator norms w.r.to the Euclidean norms on $\mathbb{R}^n$ and $\mathbb{R}^m$, that is $\|K \|_{2,2}=\|K^T\|_{2,2}$. A unit vector $f$ maximizes $\|Kf\|_2$ iff it is an eigenvector of the positive symmetric matrix $K^TK$ w.r.to its maximum eigenvalue , aka maximum singular value of $K$, squared. ( I'm not quite sure why you are then dealing with $K^T$ and not $K$, and I guess you mean $g$ to be normalized). $\endgroup$ Commented Nov 20, 2014 at 11:48

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