What can we know about the dot product of two vectors $a$ and $b$ if both $a$ and $b$ are the result of element wise multiplication?

Question

Sorry for the wordy and confusing question title, but I couldn't come up with another way to phrase it. Here is what I am talking about. With $\odot$ signifying elementwise multiplication,

$$ z = (a \odot x)^\top (y \odot b) $$

Now say that I know something about the dot product of $x^\top y$ where $x$ and $y$ are totally independent of $a$ and $b$. If I know that $x^\top y$ is going to be near zero, will that tell me anything about $z$, like perhaps $z < a^\top b$?

Answer 1

I doubt we can do something with $x^Ty$, but we can do something with $x\odot y$. Note that $$ z = (a\odot x)^T(y\odot b) = \sum_i a_ix_iy_ib_i = (a\odot b)^T(x\odot y) $$ and apply Cauchy-Schwarz to get $$ |z| \leq |a\odot b||x\odot y|. $$

Something else to note is that $|x^Ty|$ is essentially smaller than $|x\odot y|$: if $e = (1,1,1,\dotsc)$ and $n$ is our dimension, then $$ x^Ty = e^T(x\odot y) \implies |x^Ty| \leq \sqrt n|x\odot y|. $$