Proof $f(x,y)=x_1+e^{x_{2}}$ is strictly convex

Question

I am trying to show that $f(x,y)=x_1+e^{x_2}$ is strictly convex.

I can show this using the Hessian Matrix which is positive definite. However for some reason i can not put it together using algebra when it comes to two variables.

To be strictly convex then: $f(t(x_1)+(1-t)(x_2)) < t(f(x_1))+(1-t)f(x_2))$

How do i proceed?

Answer 1

This function is convex (it is the sum of two convex functions) but it is not strictly convex.

To see why, look at its behaviour on the line connecting $\mathbf{x} = (x_1, y)$ and $\mathbf{x}^\prime = (x_2, y)$. (Note that $y$ is fixed.) Then $$f(t\mathbf{x}+(1-t)\mathbf{x}^\prime) = tx_1+(1-t)x_2+e^y = tf(\mathbf{x})+(1-t)f(\mathbf{x}^\prime)$$ You get an equality that is compatible with convexity, but not the strict inequality required for strict convexity.