If $f(x)=\psi(\langle a,x \rangle )$, where $\psi:\mathbb{R} \rightarrow \mathbb{R} \in C^{2}$, then every critical point of $f$ is degenerate.

Question

If $f(x)=\psi(\langle a,x \rangle )$, where $\psi:\mathbb{R} \rightarrow \mathbb{R} \in C^{2}$, then every critical point of $f$ is degenerate. My aim here is to show that the determinant of Hessian matrix of $f$ will be 0 for every critical point of $f$. But to get there I suppose I need to find the second partial derivatives of $f$, and there is my problem. Derivating in $\mathbb{R}^n$ is very complicated to me, but I tried using the chain rule. Please verify my calculations.

$\frac{\partial}{\partial x} f(x) = \frac{\partial}{\partial x} (\psi o \langle \rangle) (a,x) = \psi ' (\langle a,x \rangle).\langle \rangle'(a,x) \Rightarrow \frac{\partial^{2}}{\partial x^{2}} f(x) = \frac{\partial}{\partial x} ( \psi'(\langle a,x \rangle)\langle \rangle'(a,x)) = \psi''(\langle a,x \rangle)\langle \rangle'(a,x)\langle \rangle'(a,x) + \psi'(\langle a,x \rangle)\langle \rangle''(a,x)$

But I can feel there is something wrong. Or is it right and should I calculate the other componentes of the Hessian?

Thanks.

Answer 1

Lets calculate the second derivatives in the case $n=2$, hopefully it get's more clear from there. First write $\langle a,x\rangle=a_1x_1+a_2x_2$. Then we have

$$f(x_1,x_2)=\psi(a_1x_1+a_2x_2) $$

Now it should be easy, we get

$$ \partial_{x_1}f(x_1,x_2)=\psi'(a_1x_1+a_2x_2)a_1 $$ $$ \partial_{x_2}f(x_1,x_2)=\psi'(a_1x_1+a_2x_2)a_2 $$

by using the chain rule. Taking this further we get

$$ Hessf(x_1,x_2)=\begin{pmatrix} \psi''(a_1x_1+a_2x_2)a_1^2 & \psi''(a_1x_1+a_2x_2)a_1a_2 \\ \psi''(a_1x_1+a_2x_2)a_1a_2 & \psi''(a_1x_1+a_2x_2)a_2^2 \end{pmatrix}= \psi''(a_1x_1+a_2x_2)\begin{pmatrix} a_1^2 & a_1a_2 \\ a_1a_2 & a_2^2 \end{pmatrix} $$

Now observe that the rows are multiples of each other, by multiplying the first row with $a_2$ and the second row with $a_1$. I leave it to you to generalize this to arbitrary dimensions. :)