Convexity of vector valued function

Suppose $$f(\vec{x}) = - \log{\sigma(\vec{x}^T \vec{w})}$$ where $$\sigma(y) = \frac{1}{1+e^{-y}}$$. Show that the function is convex.

I have computed the gradient to be $$\nabla_i f(\vec{x}) = - w_i (1+e^{-\vec{x}^T \vec{w}})$$

and the Hessian to be $$H_{ij}(\vec{x}) = \nabla^2_{ij} f(\vec{x}) = w_iw_j e^{-\vec{x}^T \vec{w}}$$

How can I argue that $$H_{ij} \geq 0$$ since I don't know the components of $$\vec{w}$$ and presumably it's possible that $$w_i>0$$ and $$w_j<0$$ which would cause a problem. Or is there an implicit assumption that $$\vec{w}>0$$ which seems unlikely?

• The matrix $v v^T$ is positive semidefinite, so the Hessian is positive semi definite so the function is convex. Sep 6, 2019 at 16:24

$$f(x)=-\log {e^y\over 1+e^y}\Big|_{y=x^Tw}=-x^Tw-\log {1\over 1+e^y}\Big|_{y=x^Tw}$$therefore we only need to show that $$-\log {1\over 1+e^y}\Big|_{y=x^Tw}$$ is convex. Now use the theorem stating that if $$f:A\to B$$ and $$g:B\to C$$ are convex, then $$fog:A\to C$$ is convex either.