Show that for every $x$ there exists a $n \in \mathbb{N}$ so that $f(x-(\frac{1}{2})^{n} \nabla f(x)) \leq f(x)-(\frac{1}{2})^{n+1} ||\nabla f(x)||^2$

Question

Let $f: \mathbb{R}^{n}\rightarrow\mathbb{R}$ be continuous and differentiable and positive. As stated in the title show that for every $x$ there exists a $n \in \mathbb{N}$ so that

$f(x-(\frac{1}{2})^{n} \nabla f(x)) \leq f(x)-(\frac{1}{2})^{n+1} ||\nabla f(x)||^2$

Given Hints: are Cauchy-Schwarz and the fundamental theorem of integral and differential calculus.

There is probably an equation of the function $f(x+y)$ so that you can rewrite it into an Integral from where I can work on, however nothing comes into my mind so far so that I could start. I hope someone will help me with this one!

Answer 1

There is a really short answer to that. Recalling that $D_x f(v)=lim_{h\rightarrow 0}\frac{f(x)-f(x+hv)}{h}=\nabla f(x)^Tv=$<$\nabla f,v$> you can rewrite the inequality.

$f(x-(\frac{1}{2})^{n} \nabla f(x)) \leq f(x)-(\frac{1}{2})^{n+1} ||\nabla f(x)||^2$

= $ (\frac{1}{2}) ||\nabla f(x)||^2 \leq \frac{f(x)-(x-(\frac{1}{2})^{n} \nabla f(x))}{\frac{1}{2}^n}$ $(1)$

Let $h=(\frac{1}{2})^{n}$ and $v=\nabla f$. We see that for $lim_{h\rightarrow 0}$ we get $\frac{1}{2}||\nabla f(x)||\leq||\nabla f(x)||^2$ in $(1)$. If we let $n$ be "big enough" the inequality therefore follows.