Why $g(t)=f(x+t(y-x))$ is non-decreasing when $\langle \nabla f(y)- \nabla f(x),y-x \rangle$?

by Saeed   Last Updated November 11, 2018 05:20 AM

Let $f:\mathbb{R}^n \rightarrow \mathbb{R}$ be a differentiable function. Now define the differentiable univariate function as follows

$$g(t)=f(x+t(y-x)), \,\,\, 0 \leq t \leq 1 $$ where $x,y \in \mathbb{R}^n$.

Show that when gradient of $f$ is monotone, that is $\langle \nabla f(y)- \nabla f(x),y-x \rangle$, $g(t)$ is non-decreasing over $0 \leq t \leq 1$.

Notice that $g'(t)=\langle \nabla f(x+t(y-x)),y-x \rangle$.

To show the statement, we need to take two points in $0\leq t \leq 1$ where $t_1 \leq t_2$, and we need to show that $g'(t_1) \leq g'(t_2)$, i.e.

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

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

$$ \langle \nabla f(x+t_2(y-x)),y-x \rangle - \langle \nabla f(x+t_1(y-x)),y-x \rangle $$ $$ = \langle \nabla f(x+t_2(y-x)) - \nabla f(x+t_1(y-x)) ,y-x \rangle \geq 0 $$

Now, how can we show the above quantity is non-negative?

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