# A simple question about an inequality of complex functions.

Let a sequence of complex functions $a_{ij} = \bar{a_{ji}} \in C^{\infty} (\Bbb R^{n+1})$ for $i,j = 1,\cdots,n$. Assume that $a_{ij} (\vec0) = 0$ for $i,j = 1, \cdots , n$. Then I want to prove that $$\exists m >0 \;\; \exists \eta >0 \;\;\; \forall \xi \in \Bbb C^n \;\; \forall v \in \Bbb C^{n+1} , \;\; |v| < \eta :$$ $$| \xi |^2 + \sum_{i,j=1}^n a_{ij} (v) \xi_i \bar\xi_j \geqslant m | \xi|^2.$$ Here bar means that $\overline{x+yi} = x-yi$ for some real $x,y$.

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## 1 Answer

Let $A(v)$ the $n\times n$ matrix whose general term is $a_{i,j}(v)$. We can find $\eta>0$ such that if $|v|\leq\eta$ then for all $i,j$, $|a_{i,j}(v)|\leq \frac 1{2n}$. By the gershgorin circles theorem, the eigenvalues of $1+A(v)$ for such $v$ are in the union of the circle of center $|1+a_{ii}(v)|$ and radius $\frac 12$. If $z$ is in such a circle, we have $|z|\geq |1+a_{ii}|-|z-(1+a_{ii})|\geq 1-\frac 1{2n}-\frac 12=\frac 12-\frac 1{2n}>0$, so we can take $m:=\frac 12-\frac 1{2n}$, using Rayleigh quotients, as we have an uniform below bound for the smallest eigenvalue of $I+A(v)$ .

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Thank you very much! I think "$A(v)$" should be replaced by "$I +A(v)$" in the second line of your answer. – Ann Oct 5 '12 at 19:07
@Ann I've corrected. Thanks! – Davide Giraudo Oct 5 '12 at 19:43