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Let $A\in{\rm End}(\Bbb K^n)$. Prove that $h:\Bbb K^n\to\Bbb K,\; x\mapsto (Ax|x)$ is continuous.

(Here $(\cdot|\cdot)$ represent an scalar product and $\Bbb K$ is $\Bbb R$ or $\Bbb C$. The notation ${\rm End}(\Bbb K^n)$ represent the set of all endomorphisms of $\Bbb K^n$.)

Known previous facts: the function $A$ is Lipschitz (any linear map in a finite space of the kind $\Bbb K^n$ is Lipschitz). Then $\|Ax\|\le K\|x\|$. And we need to prove that for all $x\in\Bbb K^n$

$$\forall\epsilon>0\exists\delta>0:\|x-y\|<\delta\implies|(Ax|x)-(Ay|y)|<\epsilon$$

First attempt: Using the Cauchy-Schwarz inequality and the Lipschitz condition in the RHS we have

$$|(Ax|x)-(Ay|y)|\le \big|\|Ax\|\|x\|-(Ay|y)\big|\le \big|K\|x\|^2-(Ay|y)\big|$$

and from the LHS using the reverse triangle inequality and squaring we have

$$K\|x-y\|^2\ge \big|\sqrt K\|x\|-\sqrt K\|y\|\big|^2=K\big|\|x\|^2+\|y\|^2-2\|x\|\|y\|\big|$$

but I dont get any idea to continue with this line of action, so I abandoned here. (If you know how to continue or develop a proof in the same spirit that this attempt I would be very happy to see it.)

Second attempt: This line of action is less explicit. I know that the function

$$f:\Bbb K^n\times\Bbb K^n\to \Bbb K,\;\langle x,y\rangle\mapsto (x|y)$$

is continuous, and any linear map as $A$ is Lipschitz what implies continuity. And the composition of continuous functions is continuous.

Then the function

$$g:\Bbb K^n\to\Bbb K^n\times\Bbb K^n,\; x\mapsto\langle Ax,x\rangle$$

is continuous. And finally the composition $g\circ f=h$ is continuous.$\Box$


Questions:

  1. There is a way to develop a proof in the spirit of attempt number one? I would like to see a more explicit proof.

  2. It is the proof in the second attempt correct?

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  • $\begingroup$ It's the composition of a bilinear function with the the identity function of $\Bbb K^n \times \Bbb K^n$. This way you just need to know that bilinear functions on finite dimensional vector spaces are continuous $\endgroup$ – Cauchy Nov 6 '16 at 22:24
  • $\begingroup$ You are right sorry $\endgroup$ – John Math Nov 6 '16 at 22:25

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