Consider the function $f:\mathbb{R}^2\rightarrow\mathbb{R}$ given by $f(x,y)=(y-x^2)(y-3x^2)$ Prove that:
If $u\in\mathbb{R^2}\setminus\left \{0\right\}$ an arbitrary vector and $g:\mathbb{R}\rightarrow\mathbb{R}^2$ given by $g(t)=tu$, then the function $f\circ g$ has a local minimun at $0\in \mathbb{R}$
I have tried to do it through the Hessian matrix but I have gotten stuck in how it should derive, I also feel that in any case I would only get 0 in the Hessian matrix, can someone help me?
On
No. need to use the Hessain matrix. $f(g(t))=(tu_2-t^{2}u_1^{2})(tu_2-3t^{2}u_1^{2})=t^{2}((u_2-tu_1^{2})(u_2-3tu_1^{2})$ which has the value $0$ when $t=0$. So we only have to check that $f(g(t)) \geq 0$ when $|t|$ is sufficiently small. But $((u_2-tu_1^{2})(u_2-3tu_1^{2}) \to u_2^{2}$ as $ t>0$ so it is clear that $((u_2-tu_1^{2})(u_2-3tu_1^{2}) >0$ when $|t|$ is sufficiently small provided $u_2 \neq 0$. The case $u_2=0$ is left to you.
If $u=(a,b)$, then $g(tu)=3 a^4 t^4-4 a^2 b t^3+b^2 t^2$, whose second derivative at $0$ is $2b^2$. So, if $b\ne0$, $g(tu)$ has a local minimum at $0$. And if $b=a$, then $g(tu)=3a^4t^4$, which also has a local minimum at $0$.