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I am trying to show that a quadratic form $Q: \mathbb{R}^n \to \mathbb{R}, Q(x)=x^T A x$ is differentiable from the definition of the differential.

I started by considering $Q(x+h)=(x+h)^T A (x+h)=x^T A x + x^T A h + h^T A x + h^T A h$.

We need to find a linear map $L_Q: \mathbb{R}^n \to \mathbb{R}$ s.t. $\lim \limits_{h \to 0} \frac{Q(x+h)-Q(x)-L_Q(h)}{\|h\|}$.

Note that $x^T (A + A^T) h = \langle x, (A + A^T) h \rangle$ is a linear map, so this is my candidate for the differential of $Q(x)$, but I am struggling to show that the error term $r_Q(x)=h^T A h$ decays sublinearly.

What I tried was to use the CS inequality to show that

$\frac{|r_Q(x)|}{\|h\|} = \frac{|\langle h, Ah\rangle|}{\|h\|} \leq \frac{\|h\| \|Ah\|}{\|h\|} = \|Ah\|$,

but I don't see how I can show that the right-hand side has a limit of 0.

Can someone please tell me if I am on the right track and give me some guidance? This is a problem in my lecture notes right after the definiton of the differential, so it should be able to solve this without much more knowdledge.

Thanks a lot!

DerivativesGuy
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1 Answers1

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You made a mistake in the second last displayed formula. In reality you have $$\bigl|\langle h,Ah\rangle\bigr|\leq|h|\>|Ah|\leq\|A\|\>|h|^2\ ,$$ where $\|A\|$ is the operator norm of $A$, and therefore $${\langle h,Ah\rangle^2\over|h|^2}\leq\|A\|^2\>|h|^2\ .$$

Christian Blatter
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  • Thanks for your answer. I am a little confused about your notation. I assume you wrote $| \cdot |$ for the Euclidean norm to distinguish it from the operator norm. Note that I did not use the operator norm since this has not been introduced at this point, but I rather cancelled the term in the numerator with the square in the denominator. What is wrong with my calculation? – DerivativesGuy Jun 06 '20 at 16:00
  • Ah, never mind. I actually did not have to square the term at all. My CS inequality is incorrect. Thanks for pointing out. – DerivativesGuy Jun 06 '20 at 16:04
  • @DerivativesGuy If this response answered your question, you should accept it. This will prevent the question from being unnecessarily bumped to the top of the page by the Community user. – Cameron Williams Sep 05 '21 at 16:18