Proof that $\|\vec{x} \odot \vec{y}\| \leq \|\vec{x}\|\|\vec{y}\|$

Question

I suspect that $$\|\vec{x} \odot \vec{y}\| \leq \|\vec{x}\|\|\vec{y}\|,$$ where $\vec{x}, \vec{y} \in \mathbb{R}^n$ and $\odot$ is the Hadamard product. I have completed Monte Carlo simulations where the components of $\vec{x}$ and $\vec{y}$ are IID standard normal distributions with dimensions 1 through 100 and found the inequality to always hold for sample sizes of $10^4$.

Of course, it is similar to the Cauchy-Schwarz inequality, but not identical as I was easily able to find examples where $\|\vec{x} \cdot \vec{y}\| \neq \|\vec{x} \odot \vec{y}\|$. There is no direct inequality between them either as one Monte Carlo simulation similar to the ones mentioned above estimated $Pr\left( \|\vec{x} \cdot \vec{y}\| \leq \|\vec{x} \odot \vec{y}\| \right) \approx 0.6733$.

Answer 1

Let $\vec x=(x_1,x_2,...,x_n)$ and $\vec y=(y_1,y_2,...,y_n)$.

Then $\|\vec{x} \odot \vec{y}\|^2=\|(x_1y_1,x_2y_2,...,x_ny_n)\|^2={x_1^2y_1^2+x_2^2y_2^2+\cdots x_n^2y_n^2}$

$\le (x_1^2+x_2^2+\cdots+x_n^2)(y_1^2+y_2^2+\cdots y_n^2)=\|x\|^2\|y\|^2$

since the cross-terms (i.e., $x_i^2y_j^2$ with $i\ne j$) are non-negative.

Answer 2

Note that we can interpret the difference between your simulations via the operator norm. For example, we have that $x\odot y = Xy$ for $X=\mathrm{diag}(x)$ so that $$ \|x\odot y\|_2 = \|Xy\|_2\leq \|X\|_2\|y\|_2 = \max_i\{|x_i|\}\|y\|_2 $$ for the operator norm on $X$.

As an aside, I wonder if the numerical result that you're getting can be explained by properties of the normal distribution and the dimension...