Hello , I have to show the following exercise included in the picture. I have managed to show the backward (from 2 to 1 ) .I simply used Taylor's theorem around y and then I applied the given inequality (2) and some linear algebra and got the (1) . Unfortunately, I am struggling with the other way (from 1 to 2). Some solutions I found online use contradiction and some other steps that I do not really understand. If someone here can help by providing a simple proof for this part I would be grateful.
To go from (1) to (2), notice that around any point $x$ we have
$$ \nabla f(y) = \nabla f(x) + \nabla^2 f(x)(y - x) + O(\|y - x\|^2). $$
Now, plug in $y = x + tv$ for an arbitrary unit vector $v$. Then
$$ \nabla f(x + tv) - \nabla f(x) = t \nabla^2 f(x) v + O(t^2). $$
Take an inner product on both sides with $v$, which gives
$$ \langle v, \nabla f(x + tv) - \nabla f(x)\rangle = t \langle v, \nabla^2 f(x) v \rangle + O(t^2). $$
Now, take absolute values on both sides and use smoothness of $f$ to upper bound the left hand side.
Can you see how to continue from there?