Let $f:\mathbb{R}^m \mapsto \mathbb{R}$ be a proper convex function that is not necessarily differentiable and let $x\in\mathbb{R}^n$ be such that $\mathbf{0} \notin \partial f(x)$. I want to prove that if $v\in\mathbb{R}^n$ is such that the directional derivative $f^\prime (x;v) < 0$ then $v$ is a descent direction of $f$ at $x$
How my proof starts: Suppose $f(x;v) <0$. Then, by convexity, we have the Taylor-like expansion $$\forall t >0: f(x + tv) \geq f(x) + f^\prime(x;tv) = f(x) + t f^\prime (x;v)$$ Rearranging terms, we have $$\frac{f(x + tv) - f(x)}{t} \geq f^\prime (x; v)$$ For ease of notation, let $\Delta_t f(x;v) := \frac{f(x + tv) - f(x)}{t}$. Then, continuing from the above expression, we have that $$\Delta_t f(x;v) \geq f^\prime (x;v) = \inf_{t>0} \Delta_t f(x;v)$$ where equality of the rightmost expression follows by convexity.
Where my proof gets shaky: Let $t^\star \in \arg\inf_{t>0} \Delta_t f(x;v)$. Then, $$\exists \varepsilon \in (0,t^\star), \forall \alpha \in [t^\star-\epsilon, t^\star + \epsilon]: \Delta_\alpha f(x;v) < 0 \implies f(x+\alpha v) < f(x)$$ which therefore means that $v$ is in fact a descent direction.
The flaws of my proof:
- The infimum is over an open set so there's no guarantee that $\arg\inf_{t>0} \Delta_t f(x;v)$ is non-empty.
- Assuming $t^\star$ does exist, my proof then depends on the fact that the difference quotient $\Delta_t f(x;v)$ is continuous in $t$ about $t^\star$, allowing me to construct a ball $[t^\star - \epsilon, t^\star + \epsilon]$. This also may not be true.
Any help is appreciated.
It seems, everything is a bit simpler. By the definition of the directional derivative we have: $$0>f'(x;v) = \lim_{t\to 0}\frac{f(x+tv)-f(x)}{t} =\inf_{t>0}\frac{f(x+tv)-f(x)}{t},$$ where the latter equality holds due to the convexity of $f$. Now by the definition of $\inf$ we got that for $t$ small enough $$f(x+tv)-f(x)\leq 0,$$ which means that $v$ is the descent direction of $f$ at $x$.