I've been working on the following exercise:

Consider constants $a, b \in \mathbb{R}$, with $a>0$. Show that the function $f:\mathbb R^2 \rightarrow \mathbb R$ given by $f(x_1, x_2)=ax_1^4+bx_2+e^{x_1^2+x^2_2}$ has a unique critical point and this is global minimum of $f$.

I tried to solve it in a practical way, taking $c= (c_1, c_2)$ a critical point and trying to show that it is unique using operations. In the end I found that $c_1=0$ and $c_2= -\displaystyle{\sqrt{\frac{W\left(\frac{b^2}{2}\right)}{2}}}$ if $b>0$, $c_2= \displaystyle{\sqrt{\frac{W\left(\frac{b^2}{2}\right)}{2}}}$ if $b<0$ and $c_2 =0$ if $b=0$, where $W$ is the product log function.

Based on this, I have two questions:

  • I know that $c_2$ is unique depending on the choice of $b$, but for this to be true $W$ must be injective, am I right?

  • It seems to me that the $f$ function is a convex function, mainly because of the graph:

enter image description here

I could prove the result by showing that each of the components of the function $f$, ie $4ax_1^4$, $bx_2$ and $e^{x_1^2+x_2^2}$ are convex, and using that the sum of convex functions is convex? (and then use this result For a strictly convex function $f:\mathbb{R}^n\to \mathbb{R}$ critical point is unique.)

  • $\begingroup$ Is the term with coefficient $b$ linear or cubic? You have it both ways right now. $\endgroup$ – Matthew Leingang Jun 10 at 19:11
  • $\begingroup$ Sorry, is linear. $\endgroup$ – PaulichenT Jun 10 at 19:12

It's not too hard to directly compute the Hessian matrix of $f$ and use the result you linked to. $$ \mathscr{H}f(x_1,x_2) = \left[\begin{matrix}2 \left(6 a x_{1}^{2} + 2 x_{1}^{2} e^{x_{1}^{2} + x_{2}^{2}} + e^{x_{1}^{2} + x_{2}^{2}}\right) & 4 x_{1} x_{2} e^{x_{1}^{2} + x_{2}^{2}}\\4 x_{1} x_{2} e^{x_{1}^{2} + x_{2}^{2}} & 2 \left(2 x_{2}^{2} + 1\right) e^{x_{1}^{2} + x_{2}^{2}}\end{matrix}\right] $$ Its determinant is: $$ \det \mathscr{H}f(x_1,x_2) = 48 a x_{1}^{2} x_{2}^{2} e^{x_{1}^{2} + x_{2}^{2}} + 24 a x_{1}^{2} e^{x_{1}^{2} + x_{2}^{2}} + 8 x_{1}^{2} e^{2 x_{1}^{2} + 2 x_{2}^{2}} + 8 x_{2}^{2} e^{2 x_{1}^{2} + 2 x_{2}^{2}} + 4 e^{2 x_{1}^{2} + 2 x_{2}^{2}} $$ This is clearly positive for all $x_1$ and $x_2$, as is the trace of $\mathscr{H}f(x_1,x_2)$ So both eigenvalues of $\mathscr{H}f(x_1,x_2)$ are positive, which means that $\mathscr{H}f(x_1,x_2)$ is positive-definite.


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