# Unique critical point and this point is global minimum of $f$.

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:

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.)

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