# Show $\frac{1}{\|A^{-1}\|}|h| \leq |Ah|$

I'm trying to understand a proof for the inverse function theorem, it is quite long and drawn out so I won't give all the details, just the bare minimum needed for where I am having difficulties.

$A$ is some invertible linear operator on $\mathbb R^n$, and $h \in \mathbb R^n$.

$\|A\|$ is operator norm, $|h|$ is euclidean norm. At one point the writer says $\frac{1}{\|A^{-1}\|} |h|\leq |Ah|$ and I don't see why that's necessarily true.

What is true is that $\|A\| \geq \frac{1}{\|A^{-1}\|}$ and so $\frac{1}{\|A^{-1}\|} |h| \leq \|A\||h|$

But $\|A\| |h| \geq |Ah|$ and not the other way around, so that does not seem like the way to prove that $|Ah|$ is larger.

Hint: $$|h| = |A^{-1}(Ah)| \le ||A^{-1}|| \; |Ah|$$
• Damn it. Should have seen that. I was sure it had something to do with $\|A\| \|A^{-1}\| \geq 1$. Wrong direction I guess. – Oria Gruber Mar 8 '18 at 18:56
• @OriaGruber The skill is the same, but you don't have to pull $A$ out of $|Ah|$, which is present in the desired inequality. What you've done in the last paragraph introduces a bigger term $||A|| |h|$, and this leads you away from the result. – GNUSupporter 8964民主女神 地下教會 Mar 8 '18 at 18:59
You have $$|h|=|A^{-1}Ah|\leq\|A^{-1}\|\,|Ah|.$$ This inequality holds for any invertible operator on a normed space.