Let $K$ be a real valued function such that $\int_{\mathbb{R}} |K|=1$ and $||K|| < +\infty$ where $||.||$ is the $L^2$ norm which is derived from the scalar product $<f,g> =\int _{\mathbb{R}} fg$.
Assume that for all $0 \leq x \leq 1$, $$ \frac{<K,K(x.)>}{ \left\| K\right\|^2}\geq 1,$$ where $K(x.)$ is the function defined by $K(x.)(y)=K(xy)$.
Let's also define $K_h:x\mapsto \frac{1}{h}K\left ( \frac{x}{h} \right )$ for $h> 0$. I believe we also have $h \leq 1$ but i'm not sure, the way $h$ is defined on the paper this result comes from is unclear. I'm struggling to understand why the above inequality imply that for all $h' \leq h$, $$\left\| K_{h'}-K_{h}\right\|^2 \leq \left\| K_{h'} \right\|^2 - \left\| K_{h}\right\|^2.$$
I have noticed that $\left\| K\right\|^2=h \left\| K_{h}\right\|^2= h'\left\| K_{h'}\right\|^2$ but I was unsuccesful in using that equality to get what I want.
First notice that $$ \left\lVert K_{h'}-K_{h}\right\rVert^2+\left\lVert K_{h}\right\rVert^2 =2\left\lVert K_{h}\right\rVert^2+\left\lVert K_{h'}\right\rVert^2 -2\langle K_{h'},K_h\rangle.$$ Then the inner product term can be controlled by expressing it as an integral over $y$ and making the substitution $t=y/{h'}$ and using the assumption $\frac{\langle K,K(h'/h.)\rangle}{ \left\| K\right\|^2}\geq 1$.