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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

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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 so I'm putting an answer here to mark it as solved.

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