Convexity of logarithmic function with inner product and $\ell_1$-norm

Question

Consider vectors $\mathbf{x}\in\mathbb{R}^n$ s.t. $$\mathbf{x}= \begin{bmatrix} \mathbf{w}\\b \end{bmatrix}$$ where $\mathbf{w}\in\mathbb{R}^{n-1}$ and $b\in\mathbb{R}$. I want to show that the following function is convex in $\mathbf{x}$: $$f(\mathbf{x})=\mathrm{log}\left(1+e^{-\mathbf{w}^T\mathbf{y}-b+\epsilon\parallel\mathbf{w}\parallel_1}\right)$$ for fixed $\mathbf{y}\in\mathbb{R}^{n-1}$ and $\epsilon>0$. The domain $\mathrm{dom}f$ is \begin{eqnarray*} \mathrm{dom}f&=&\left\{\mathbf{x}\in\mathbb{R}^n | 1+e^{-\mathbf{w}^T\mathbf{y}-b+\epsilon\parallel\mathbf{w}\parallel_1}>0\right\}\\ &=&\mathbb{R}^n \end{eqnarray*} which is affine and hence convex. I have computed the Hessian of the function inside the logarithm, i.e., that of $$g(\mathbf{x})=1+e^{-\mathbf{w}^T\mathbf{y}-b+\epsilon\parallel\mathbf{w}\parallel_1}$$ w.r.t. to $\mathrm{x}$. However, it has some ugly expressions and I find it hard to see if it is positive definite. Since the logarithm function is concave I am wondering how the composition rules can be applied here (perhaps $f(\mathbf{x})$ is concave). Any thoughts on how to approach this?

Answer 1

Consider the function $f (t)=\log(1+e^{t})$. This is increasing and it is also convex: $f'(t)=\frac {e^{t}} {1+e^{t}}$ which gives $f''(t)=\frac {e^{t}} {(1+e^{t})^{2}} >0$. The function $-w^{T}y-b+\epsilon\|w\|_1$ is easily verified (from definition) to be a convex function. Since an increasing convex function of a convex function is convex the proof is complete.