Suppose $n,m\geq 1$, $A\subset \mathbb{R}^m$ is compact and $f:\mathbb{R}^n\times A\to \mathbb{R}^n$ is a continuous function satisfying $$\left\lVert f(x_1,a)-f(x_2,a)\right\rVert \leq L\left\lVert x_1-x_2\right\rVert \qquad (L>0)$$ for all $x_1,x_2 \in \mathbb{R}^n$ and $a\in A$.
I have to prove that $$f(y, a)\cdot y\leq K(1+{\left\lVert y\right\rVert}^2)\qquad [1]$$ for all $y\in \mathbb{R}^n$ and $a\in A$ where $$K=L+\sup_{a\in A}{\left\lVert f(0,a)\right\rVert}$$ and $\cdot$ is the scalar product on $\mathbb{R}^n$.
For all $a \in A$ and $y \in \mathbb{R}^n$ we have, by the Cauchy-Schwarz inequality and the Lipschitz condition on $f$, $$[f(y,a)-f(0,a)]\cdot y\leq \vert[f(y,a)-f(0,a)]\cdot y|\leq \left\lVert f(y,a)-f(0,a)\right\rVert\left\lVert y\right\rVert\leq L{\left\lVert y\right\rVert}^2,$$ and therefore $$f(y,a)\cdot y\leq f(0,a)\cdot y +L{\left\lVert y\right\rVert}^2\leq\left\lVert f(0,a)\right\rVert\left\lVert y\right\rVert+L{\left\lVert y\right\rVert}^2 \\ \leq \sup_{a \in A}\left\lVert f(0,a)\right\rVert \left\lVert y\right\rVert +L{\left\lVert y\right\rVert}^2. $$ Now the author concludes saying that this implies $[1]$, but I really don't understand why.
Any hint?
Thanks a lot in advance
When $\left\lVert y\right\rVert\ge 1$, we find that $$ \sup_{a \in A}\left\lVert f(0,a)\right\rVert \left\lVert y\right\rVert +L{\left\lVert y\right\rVert}^2\le K\|y\|^2\le K(\|y\|^2+1). $$ When $\left\lVert y\right\rVert\le 1$, we can observe $$ \sup_{a \in A}\left\lVert f(0,a)\right\rVert \left\lVert y\right\rVert +L{\left\lVert y\right\rVert}^2\le K\le K(\|y\|^2+1). $$ Thus we have $$ \sup_{a \in A}\left\lVert f(0,a)\right\rVert \left\lVert y\right\rVert +L{\left\lVert y\right\rVert}^2\le K(\|y\|^2+1). $$ as desired.