Hessian Related convex optimization question

Question

My precise question is from an exercise;

Let $f : \mathbb{R}^2 → \mathbb{R}$ be a twice differentiable function. Prove that there exists a $λ ∈ R$ such that $g : \mathbb{R}^2 → \mathbb{R}$ defined as $g(x) := f (x) + (λ/2)||x||^2$ is convex.

Hint: compute the Hessian of $g$ and prove that for $λ$ large enough, this Hessian is positive definite.

I am trying to follow the hint but immediately running into errors when computing the hessian of g.

I am really not sure how to proceed given the $f(x)$ along with $||x||$. I mean we only still have the one variable $x$, so what possibility is there for double derivatives?

$$ \left[ \begin{array}{ c c } g_{ff} & g_{fx} \\ g_{fx} & g_{xx} \end{array} \right] $$

Which seems rather silly. Would anyone care point out my misinterpration/fundamentally flaw, and lead me on the right track?

Answer 1

By considering $\| \cdot\|$ the $2$-norm, the Hessian is given by

\begin{equation} H = \left [ \begin{array}{cc} \frac{\partial^2 f(x_1,x_2)}{\partial x_1^2} + \lambda & \frac{\partial^2 f(x_1,x_2)}{\partial x_1 \partial x_2} \\ \frac{\partial^2 f(x_1,x_2)}{\partial x_2 \partial x_1}& \frac{\partial^2 f(x_1,x_2)}{\partial x_2^2} + \lambda\\ \end{array} \right], \end{equation} for $x = (x_1,x_2)$.

The matrix $H$ is positive definite if and only if all principal minors are positive. Thus,

$$ \frac{\partial^2 f(x_1,x_2)}{\partial x_1^2} + \lambda > 0,$$ and

$$\left( \frac{\partial^2 f(x_1,x_2)}{\partial x_1^2} + \lambda \right)\left( \frac{\partial^2 f(x_1,x_2)}{\partial x_2^2} + \lambda \right) - \left(\frac{\partial^2 f(x_1,x_2)}{\partial x_1 \partial x_2} \right)\left(\frac{\partial^2 f(x_1,x_2)}{\partial x_2 \partial x_1} \right) >0.$$

You have to assume other constraints for $f(x)$. Notice that you should have $$ \frac{\partial^2 f(x_1,x_2)}{\partial x_1^2} > -\infty,$$ for all $x_1$ and $x_2$, otherwise there is no $\lambda$ that guarantees the convexity of $g(x)$. I take the example given by @SZhu as follows.

If $\frac{\partial^2 f(x_1,x_2)}{\partial x_1^2}=-e^{x_1+x_2}$, then there is no fixed $\lambda$ such that the first principal minor is positive for all $x_1$ and $x_2$.

Yet, you also should have

$$ \frac{\partial^2 f(x_1,x_2)}{\partial x_2^2} > -\infty$$ and

$$\left(\frac{\partial^2 f(x_1,x_2)}{\partial x_1 \partial x_2} \right)\left(\frac{\partial^2 f(x_1,x_2)}{\partial x_2 \partial x_1} \right) < \infty,$$ for all $x_1$ and $x_2$.

If these conditions are satisfied then for $\lambda$ large enough, it is easy to see that the two principal minors are positive.