# directional derivative sublinear of a convex function sublinearity problem to show

How to show the following:

If $f:\mathbb R^d \rightarrow \mathbb R$ is convex

then its directional derivative is sublinear?

Thank you...

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I assume that you mean that the directional derivative is sublinear with respect to the direction? In that case we must prove that

1. $f'(x;\alpha t) = \alpha f'(x; t)$
2. $f'(x;t + s) \leq f'(x;t) + f'(x;s)$

Proof:

1. $$f'(x;\alpha t) = \lim_{h \rightarrow 0 } \frac{f(x+h\alpha t) - f(x)}{h} = \alpha \lim_{h \rightarrow 0 } \frac{f(x+(h\alpha) t) - f(x)}{h\alpha} = \alpha f'(x;t)$$

2. $$f'(x;t + s) = \lim_{h \rightarrow 0 } \frac{f(x+h(t+s)) - f(x)}{h} \\ \\= \lim_{h \rightarrow 0 } \frac{f(\frac{1}{2}(x+2ht) + \frac{1}{2}(x+2hs)) - \frac{1}{2}f(x) - \frac{1}{2}f(x)}{h} \\ \leq \lim_{h \rightarrow 0 } \frac{f((x+2ht)) - f(x)}{2h} + \lim_{h \rightarrow 0 } \frac{f((x+2hs)) - f(x)}{2h} \\ = f'(x;t) + f'(x;s)$$

In the inequality I used the convexity of $f$. Don't hesitate to ask if something i unclear. Best regards

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Thank you for the solution! –  Salih Ucan Feb 14 '13 at 23:50