Why $(y-\overline y)^t[f(x)-f(\overline x)]\ge-\Vert y-\overline y\Vert\Vert f(x)-f(\overline x)\Vert$?

228 Views Asked by At

Let $f:\mathbb R^n\to\mathbb R^{m+l}$ be continuous and $\overline y\in\mathbb R^{m+l}$.

Why $0\ge (y-\overline y)^t[f(x)-f(\overline x)]\ge-\Vert y-\overline y\Vert\Vert f(x)-f(\overline x)\Vert$ ?

What relation exists between the transpose of a vector and the norm with minus sign?

Thank you.


Edit: sorry I forgot the 0 at the beginning of the inequality.

1

There are 1 best solutions below

0
On BEST ANSWER

We have claim that $$ v\cdot w\geq -|v||w| $$

Consider $f: \mathbb{R}^n\rightarrow \mathbb{R},\ f(v)=|v|$.

For $|v|=R,\ |w|=r$, then $|f(v)-f(w)|=|R-r|$. Clearly, $$ |v-w|\geq |R-r|$$

Hence $$ ||v|-|w||^2\leq |v-w|^2 $$

Hence $$|v||w|\geq v\cdot w $$