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Let $f:\mathbb{R}^2\to\mathbb{R}$ be continuously differentiable $\ s.t\ \ \forall (x,y)\in\mathbb{R}^2 \ 2f(x,y)=f(2x,4y)$

Prove that there exists $c\in\mathbb{R} \ s.t \ f(x,y)=cx \ \forall (x,y)\in\mathbb{R}^2$

My work:

My motivation was to use the chain rule, so I defined $g:\mathbb{R}^2\to\mathbb{R}^2 \ by \ g(x,y)=(2x,4y)$

Now by what was given to us, we deduce for arbitrary $x_0,y_0$:

$(2f_x,2f_y)|_{(x_0,y_0)}=2Df(x,y)|_{(x_0,y_0)}=Df(2x,4y)|_{(x_0,y_0)}=D(f\circ g(x,y))|_{(x_0,y_0)}$

$==<(f_x,f_y)|_{(2x_0,4y_0)},\begin{pmatrix} 2 &0 \\ 0& 4 \end{pmatrix}>=(2f_x,4f_y)|_{(2x_0,4y_0)}$

Now I'm stuck, can anyone help?

1 Answers 1

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HINT

The essential point is that $f(x,y)$ is continuously differentiable, and if your proof does not use that fact, then it must be incorrect.

Consider $$f(x,y) = (xy)^{\frac13}$$ For this function, $f(2x,4y) = 2^{\frac13} 4^{\frac13} (xy)^{\frac13}$$ = 2f(x,y)$ yet it cannot be expressed as $f(x,y)=cx$. The point is that this is not continuously differentiable.

So the trick in the proof will be to consider a region arbitrarily close to the origin, show that the form of $f(x,y)$ necessarily goes to $$ f(x,y) = k x^a y^b $$ with $a+2b = 1$, then show that either $a$ or $b$ must be less than $\frac13$ which implies that the function is not continuously differentiable at the origin.

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    Thank you for answering! Using the chain rule though does use the fact that f is continuously differentiable, no?... Also, I have no clue on how to prove $f(x,y)=kx^ay^b$. Is there any other hint or alternative answer you can share? – 2017-01-30