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Suppose $f: \mathbb{R}^n \rightarrow \mathbb{R}$ is a differentiable function. Define $g(t)=f(x+t(y-x))$ for all $x, y \in \mathbb{R}^n$ and $t\geq 0$. Show when $\langle \nabla f(y) - \nabla f(x), y-x \rangle \geq 0$, $g'(t)=\langle \nabla f(x + t(y-x)),y-x \rangle$ is an increasing function where $g'$ is the directional derivative of function $f$ along $y-x$.

My try:

Since $f$ is differentiable, using chain rule we can write $g'(t)=\langle \nabla f(x + t(y-x)),y-x \rangle$. Suppose $t_2>t_1$, we should show $g(t_2)\geq g(t_1)$.

Saeed
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1 Answers1

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Define the following:

$g'(t_2)=\langle \nabla f(x + t_2(y-x)),y-x \rangle$

$g'(t_1)=\langle \nabla f(x + t_1(y-x)),y-x \rangle$

$g'(t_2)-g'(t_1)=\langle \nabla f(x + t_2(y-x)) -\nabla f(x + t_1(y-x)),y-x \rangle$

Suppose $t_2>t_1$. Then, you can write:

$g'(t_2)-g'(t_1)=\frac{1}{t_2-t_1}\langle \nabla f(x + t_2(y-x)) -\nabla f(x + t_1(y-x)),(t_2-t_1)(y-x) \rangle$

$=\frac{1}{t_2-t_1}\langle \nabla f(x + t_2(y-x)) -\nabla f(x + t_1(y-x)),x-x+(t_2-t_1)(y-x) \rangle$

$=\frac{1}{t_2-t_1}\langle \nabla f(x + t_2(y-x)) -\nabla f(x + t_1(y-x)),x+t_2(y-x)-(x+t_1(y-x)) \rangle $

Since $\langle \nabla f(y) - \nabla f(x), y-x \rangle \geq 0$ for all $x, y$, you get $g'(t_2)-g'(t_1)\geq 0$.