# How to use the convexity of the norm with the limit?

I have a function $g$ which is simply the norm: $g(x)=||F(x)||$. $F(x)$ is continuously differentiable. In fact it an actually be any norm and these results should still hold since all norms are convex. I have the following two results, and I am trying to figure out how to connect them to a third result:

1) I know that all norms are convex. To remind us, convexity means $\theta f(x)+(1-\theta) f(y) \ge f(\theta x+(1-\theta )y)$ so for the norm that means $\theta ||x||+(1-\theta) ||y|| \ge ||(\theta x+(1-\theta )y||$ where $\theta \in \{0,1\}$.

Convexity also means that the first order approximation is always an underestimator: $f(y) \ge f(x) + \nabla f^T*(y-x)$ which in our case by subustituting what $g(x)$ is, means $g'(x;d) \le ||F(x+hd)|| - ||F(x)||$.

2) I have the following result for the directional derivative at $g$ (the derivative of $g(x)$ in the direction $d$ : $g'(x;d)=\lim_{h\rightarrow0} \frac{||F(x)+hF'(x)d||-||F(x)||}{h}$

And I need to be able to obtain the result:

$\,\,\,\,\,\,g'(x;d) \le ||F(x)+F'(x)d||-||F(x)||$.

what i tried

I was able to use the fact that $g$ is convex, and $g'(x;d) \le ||F(x+hd)|| - ||F(x)||$, and substitute that in to show

$\lim_{h\downarrow0} \frac{||F(x)+hF'(x)d||-||F(x)||}{h} \le ||F(x+hd)||-||F(x)||$ But I cannot seem to obtain the result I need. Can someone help me obtain the result that I need?

For fixed $x$ and $d$, let $$\phi(t) = \left\|F(x) + tF'(x)d\right\|$$ By convexity of the norm $\left\|\cdot\right\|$, $\phi$ is also convex and so for $0\le t \le 1$, $$\phi(t) = \phi((1-t)\cdot 0 + t\cdot 1) \le t\phi(1) + (1-t)\phi(0)$$ Implies $$\frac{\phi(t) - \phi(0)}{t} \le \phi(1) - \phi(0)$$ $t\rightarrow 0$ and you find what you are looking for.