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Consider $f:\mathbb{R}^2\to\mathbb{R}\; f(x,y)=(x^2-y)(x^2-3y)$ and a linear function $g:\mathbb{R}\to\mathbb{R}^2,\; x\mapsto \begin{pmatrix} g_x(x)\\ g_y(x) \end{pmatrix} $. The claim is: If $g$ is not constant zero, then $f\circ g:\mathbb{R}\to\mathbb{R}$ has a local minimum at zero.

My try: f and g are differentiable, therefore $f\circ g$ is differentiable and $f'(g_x(x),g_y(x))g'(x)=4g_x^3(x)g_x'(x)-8g_x(x)g_y(x)g_x'(x)-4g_x^2(x)g_y'(x)+6g_y(x)g_y'(x)$ by chain rule.
Now, one solution of $f'(g_x(x),g_y(x))g'(x)=0$ is $x=0$ because g is linear and therefore it is $g^n(0)=g^{n-1}(0)g(0)=0$ for every $n\in\mathbb{N}$. This means, $x=0$ is a critical point of $f\circ g$. Let $g$ be nonzero. How to prove that $x=0$ is a local minimum of $f\circ g$? Shall I calculate the second derivation of $f\circ g$? Is there any possibility to calculate this without much computation? Mabye it could be helpful to write $g_x(x)=ax+b$ and $g_y(x)=cx+d$ with $a,b,c,d\in\mathbb{R}$ and either $a$ or $b$ is nonzero, but it could be an overkill.

But the derivation of a given linear transform is well-known, see for example here Derivative of a linear transformation.. Maybe it is helpful if you want to calculate the derivatives. But I don't know how to continue exactly.

Could you help me how to continue? Regards

Community
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dambinub
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1 Answers1

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Suppose $h=f\circ g , h(x) = (g_x^2(x)-g_y(x))(g_x^2(x)-3g_y(x))$

Then $h''(x) = 12g_x^2(x){g'}_x^2(x)-8{g'}_x^2(x)g_y(x)-16g_x(x)g'_x(x)g'_y(x)+6{g'}_y^2(x)$

(Since $g$ is linear , $g''_x = g''_y=0$)

As you suppose ${g_x}(x)=ax , {g_y}(x)=cx$

Conclude that $h''(x) = 12a^2(ax)^2-8a^2(cx)-16(ax)ac+6c^2$

$h''(0)=6c^2>0$

With $h'(0)=0 $ as you get.

Zero is a local minimum.

Brian
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  • thank you. But g could be constant too, I have to consider all cases, right? – dambinub May 24 '15 at 13:06
  • @dambinub Oh,yes,I just supposed g is a linear map. You may deal with it more precisely to show $h'(0)=0 ,h''(0)>0$ – Brian May 24 '15 at 13:13
  • Ok, thank you! I'm stuck with $h''(0)>0$. Because if $g_x(x)=ax+b$ and $g_y(x)=cx+d$, a,b,c,d are real numbers and at least one of these numbers is nonzero, I obtain: $h''(0)=12b^2a^2-8a^2d-16bac+6c^2$. But I have no idea why $h''(0)>0$. – dambinub May 24 '15 at 13:29
  • oh no, everything is correct. $g_x$ and $g_y$ do not have the form $g_x(x)=ax+b$ and $g_y(x)=cx+d$. You are right to consider $g_x(x)=ax$ and $g_y(x)=cx$. Thank you!! – dambinub May 24 '15 at 14:14