# Is this proof about homogeneous functions valid?

Let $$f:\mathbb{R}^2\to\mathbb{R}$$ be homogeneous and suppose that $$f(a, b)=0$$ for all $$(a, b)$$ where $$a^2+b^2=1$$. Show that $$f(x, y)=0$$ for all $$(x, y) \neq (0,0)$$.

If $$f$$ is homogeneous, then it can be written as $$f(tx, ty)=t^{\lambda}f(x, y)$$ Since $$\left(\frac{x}{\sqrt{x^2+y^2}}\right)^2+\left(\frac{y}{\sqrt{x^2+y^2}}\right)^2 = 1$$, then $$f\left(tx,ty\right)=(\sqrt{x^2+y^2})^{\lambda}f\left(\frac{x}{\sqrt{x^2+y^2}},\frac{y}{\sqrt{x^2+y^2}}\right)=0$$

Is this proof valid?

• You mean take $t=\sqrt{x^2+y^2}$. It seems right. – Tito Eliatron Nov 8 '18 at 20:36

Let $$(a,b)$$ be such that $$(a,b)\not=(0,0)$$. Let $$t=\frac{1}{\sqrt{a^{2}+b^{2}}}$$. Note that $$(ta)^{2}+(tb)^{2}=\frac{a^{2}}{a^{2}+b^{2}}+\frac{b^{2}}{a^{2}+b^{2}}=1.$$ Hence, $$t^{\lambda}f(a,b)=f(ta,tb)=0,$$ by assumption. Hence, $$f(a,b)=0$$.