Let $k(.,.)$ be a function that takes two vectors as input and outputs a scalar as follows
\begin{align}
\mathcal{k}(x,y) = \exp(-\frac{||x-y||\_2^2}{2})
\end{align}
where $||x||\_2$ denotes the $2-$norm. Now, let $x\_1,\dots,x\_m$ be $m$ vectors in $\mathbb{R}^d$. Let me define the $m \times m$ matrix $\mathbf{K}$ such that its entries are given as
\begin{align}
\mathbf{K}\_{ij} = \mathcal{k}(x\_i,x\_j)
\end{align}
For any vector $x$, let me define the $m\times 1$ vector $\mathcal{K}\_x$ as 
\begin{align}
\mathcal{K}\_x = \begin{bmatrix}
\mathcal{k}(x,x\_1) \\ \vdots \\ \mathcal{k}(x,x\_m)
\end{bmatrix}
\end{align}
Now I am curious if following inequality holds true for every $x\in\mathbb{R}^d$
\begin{align}
\mathcal{K}\_x^T\mathbf{K}^{-1}\mathcal{K}\_x \leq 1
\end{align}



Inequality on numerical range of inverse of kernel matrix
The *radial basis function kernel* $\mathcal{k}(x,y) = \exp(-\frac{||x-y||\_2^2}{2\sigma^2})$ is a *positive definite kernel*. So,
$M=\begin{bmatrix}\textbf{K}&\mathcal{K}\_x\\\mathcal{K}\_x^T&1\end{bmatrix} $
is a positive definie matrix. Hence, the *Schur complement* of the block $1$ of the matrix $M$ is also a positive definie matrix. That is $1-\mathcal{K}\_x^T\mathbf{K}^{-1}\mathcal{K}\_x \geq 0$. Q.E.D.


