# 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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