It seems intuitive that given a $ C^1 $ function $ f : \mathbb R \to \mathbb R $, where
- for some $ t_0 > 0 $ we have $ |f(x+t)| < |f(x)| $ for all $ t \in (0,t_0) $, and
- $ f'(x) \neq 0 $,
there must be $ t_1 \in (0,t_0) $ such that $ |f(x)+f'(x)t| < |f(x)| $ as well for $ t \in (0,t_1) $.
I just can't seem to figure out how to prove this. Is my intuition completely wrong here?
If it is not wrong, does the idea generalize to arbitrary Banach spaces $ V,W $ and a Fréchet differentiable map $ G : V \to W $?
That is, with fixed $ x,y \in V $ and $ G $ satisfying
- for some $ t_0 > 0 $ we have $ \|G(x+ty)\|_W < \|G(x)\|_W $ for all $ t \in (0,t_0) $, and
- $ G'(x)[y] \neq 0 $ in $ W $.
Is there $ t_1 \in (0,t_0) $ such that $ \|G(x)+tG'(x)[y]\|_W < \|G(x)\|_W $?
It seems to me that where this could go wrong is if $ \|G(x)\|_W-\|G(x+ty)\|_W \to 0 $ as $ t \to 0 $ faster than $ \mathcal O(t) $. But I can't seem to figure out if this could be possible?
Update:
Observations - For the second part, there seem to be special scenarios where his breaks down at least. If say $ G'(x)[y] $ is in the direction of an isoline of $ \|\cdot\|_W $, things could break down.
Example - Let $ V,W = \mathbb R^2 $, but with 1-norm. Let $ G(x) = x - x^2 $, where multiplication is considered elementwise. Then
$$ G(x+ty) = G(x) + t\underbrace{y(1-2x)}_{G'(x)[y]} - t^2y^2, $$
and (though not proven here) this $ G $ is Fréchet differentiable.
If we assume $ 0 < x_i < \frac12 $, then $ G(x)_i > 0 $ and $ (1-2x_i) > 0 $. Pick $ y_1 = -(1-2x_1)^{-1} $ and $ y_2 = (1-2x_2)^{-1} $. Now, assuming $ t $ sufficiently small,
\begin{align*}
\|G(x)\|_1 &= (x_1-x_1^2) + (x_2 - x_2^2) \\
\|G(x) + tG'(x)[y]\|_1 &= (x_1-x_1^2-t) + (x_2-x_2^2+t) = \|G(x)\|_1 \\
\|G(x+ty)\|_1 &= (x_1-x_1^2-t-t^2y_1^2) + (x_2-x_2^2+t-t^2y_2^2) \\
&= (x_1-x_1^2-t^2y_1^2) + (x_2-x_2^2-t^2y_2^2) = \|G(x)-t^2y^2\|_1 < \|G(x)\|_1
\end{align*}
Thus we have a contradiction. QED
What kind of assumptions could be added to remedy this? Is assuming $ G(x)[y] $ to not point along an isoline of $ \|\cdot\|_W $ enough? What if the strict inequalities where replaced with "less than or equal"?