# $\displaystyle\sum_{n=0}^{\infty}(f(x_n,y_n))^2$ converges.

Let $f:\mathbb{R}^2\rightarrow \mathbb{R}$ be a continuous mapping such that: $\forall (\lambda ,x,y)\in\mathbb{R}^3,f(\lambda x,\lambda y)=\lambda f(x,y)$.

Let $(x_n)_{n\geq0}$,$(y_n)_{n\geq0}$ be two real sequences such that $\displaystyle\sum_{n=0}^{\infty}x_n^2$ and $\displaystyle\sum_{n=0}^{\infty}y_n^2$ coverge.

Prove that: $\displaystyle\sum_{n=0}^{\infty}(f(x_n,y_n))^2$ coverges.

I've intended to find $m\in\mathbb{R}_+:f(x,y)\leq m \ \forall x\leq 1,y \leq 1$.

Since $\forall(x,y)\in \mathbb{R}^2-\left \{ (0;0) \right \}$,

$\left | f(x;y) \right |=\sqrt{x^2+y^2}\left | f(\frac{x}{\sqrt{x^2+y^2}};\frac{y}{\sqrt{x^2+y^2}}) \right |\leq m\sqrt{x^2+y^2}$

$f(0;0)=0$. Hence, $\forall(x,y)\in \mathbb{R}^2$, $\left | f(x;y) \right |\leq m\sqrt{x^2+y^2}\Rightarrow (f(x;y))^2\leq m^2(x^2+y^2)$

Since $\sum x_n^2$ and $\sum y_n^2$ coverge, we have $\sum m^2(x_n^2+y_n^2)$ converges. Hence $\displaystyle\sum_{n=0}^{\infty}(f(x_n,y_n))^2$ coverges.

Could you help me to find $m$ or give me other solution?

Without loss of generality, suppose $x_n$ and $y_n$ are nonzero for every $n$. Note that $f(0,0) = 0$; since $f$ is continuous, there is a $\delta > 0$ such that $|f(x,y)| < 1$ whenever $\|(x,y)\| < \delta$. Set $u_n = x_n/\|(x_n,y_n)\|$ and $v_n = y_n/\|(x_n,y_n)\|$, for $n = 1,2,3,\ldots$. The condition on $f$ implies $$f^2(x_n,y_n) = \delta^{-2}(x_n^2 + y_n^2)f^2(u_n \delta, v_n \delta) \le \delta^{-2}(x_n^2 + y_n^2)$$ The result now follows from the comparison test.