On page 89 of Boyd & Vandenberghe's Convex Optimization, it says if $f: \mathbb{R}^n \to \mathbb{R}$ is convex, then its perspective $g: \mathbb{R}^{n+1}\to \mathbb{R}$ defined by
$$g(x,t) = t \, f \left( \frac{x}{t} \right)$$
where $\textbf{dom} \, g = \left\{ (x,t) \mid \frac{x}{t} \in \textbf{dom}\ f, t > 0 \right\}$, is still convex.
I'm wondering whether the converse still holds, i.e., $f(x)$ is convex if $g(x,t)$ is convex. Intuitively, I think this is right but I'm not sure since I haven't seen any materials mentioned this.
The authors proved the primal result by using epigraph and perspective mapping below. And since both inverse image and image of perspective mapping perserve convexity. Can we use the same method to prove that the converse also holds?
Best Answer
If $g$ is convex, then $$ f(\alpha x+(1-\alpha)y) = g(\alpha x+(1-\alpha)y , 1) \leq \alpha g(x , 1) + (1-\alpha) g(y,1) = \alpha f(x) + (1-\alpha)f(y) $$ and therefore, $f$ is convex.