# Continuously Differentiable and Stable; Can't remove term from solution.

Let $F:\mathbb{R}^3 \to \mathbb{R}^3$ be continuously diﬀerentiable and stable. Prove that for any $x ∈ \mathbb{R}^3$ and for any $h ∈ \mathbb{R}^3$, $||DF(x)h|| ≥ ||h||$.

I'm given that $F$ is stable so I know $\exists c>0$ such that $||F(u) - F(v)|| \geq c||u-v||$. Then by the first-order approximation theorem $$\lim_{h\to 0} \frac{||F(x+h) - F(x) - DF(x)h||}{||h||} =0$$ Where $DF(x)$ is the derivative matrix of the function $F$ at $x$. By the first-order approximation $\exists r > 0$ such that $||h|| < r$ gives $$\frac{||F(x+h) - F(x) - DF(x)h||}{||h||} <\frac{c}{2}$$ Then for $||h|| < r$ we get $$c||h|| \leq ||F(x+h) - F(x)||= ||F(x+h) - F(x) - DF(x)h + DF(x)h||$$ and $$||F(x+h) - F(x) - DF(x)h + DF(x)h|| \leq ||F(x+h) - F(x) - DF(x)h||+|| DF(x)h||$$ and lastly $$||F(x+h) - F(x) - DF(x)h||+|| DF(x)h|| < \frac{c||h||}{2} + ||DF(x)h||$$ which yields $||DF(x)h|| ≥ \frac{c||h||}{2}$.

The problem is how to get rid of the $\frac{c}{2}$ term on the right hand side. I don't seem to have any ability to pick $c$ by the theorems used and I can't say anything about its size compared to $||h||$ or $1$ so I can't justify removing it.

Any hints that anybody can provide would be appreciated.

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What is the definition of stability? Stable in what sense? If it is just the condition you wrote down, $F(x) =\frac12 x$ would be a counterexample to the claim. – Lukas Geyer Oct 22 '12 at 13:30
Yeah, It turns out there was a typo on the assignment and it was not true as it was written. Not sure how to close this without writing an answer. – AvatarOfChronos Oct 22 '12 at 19:05