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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?

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  • $\begingroup$ You mean take $t=\sqrt{x^2+y^2}$. It seems right. $\endgroup$ – Tito Eliatron Nov 8 '18 at 20:36
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You have the right idea, but I would add some more steps for comprehensibility.

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, $f(a,b)=0$.

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