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Let $\mathbf{A}(t):\mathbb{R}\rightarrow\mathbb{C}^{n\times n}$ be a rank deficient periodic function-valued positive semi-definite Hermitian matrix. The entries $a_{ij}(t)$ of $\mathbf{A}(t)$ are analytic functions. Now define $\mathbf{B}(t)=\lambda(t)\mathbf{x}(t)\mathbf{x}(t)^H$, where $\lambda(t)$ is the largest eigenvalue of $\mathbf{A}(t)$ (it is simple for all $t\in\mathbb{R}$) and $\mathbf{x}(t)$ is its associated normalized eigenvector ($()^H$ is the hermitian transpose). Do the Fourier series of the entries $b_{ij}(t)$ of $\mathbf{B}(t)$ converge pointwise to $b_{ij}(t)$?

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  • $\begingroup$ Thanks for the response, I forgot to mention that the matrix is positive semi-definite. I have added it to the definition. $\endgroup$ – trienko Dec 5 '13 at 7:18
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    $\begingroup$ A couple questions: (a) Why did you mark "rank deficient" in bold? I don't see how this changes the question much, (b) Can you assume the largest two eigenvalues of $A(t)$ are unequal for all $t$? If not, then $x(t)$ becomes ambiguous and may even move discontinuously. My initial answer to the question would be of course the Fourier series converge pointwise - I think $\lambda(t)$ and $x(t)$ are smooth functions of $t$. $\endgroup$ – Will Nelson Dec 16 '13 at 4:54
  • $\begingroup$ @WillNelson I think "the largest eigenvalue is simple for all $t\in \mathbb R$" answers your question (b) in affirmative. I also don't think that rank matters. Yes, the functions in question should be smooth (based on the spectral gap being bounded away from $0$), but somehow I could not put together a neat proof -- hence the bounty. $\endgroup$ – Post No Bulls Dec 17 '13 at 2:42
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The answer is yes! In fact $B$ is real analytic. (Not holomorphic, as you normalize.) Thus the Fourier series converges in any sense you want. Rank deficiency is irrelevant, as you can add a suitable multiple of the identity matrix, and pretty much nothing changes. The proof is not trivial. See Kato's Perturbation Theory.

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Yes, the Fourier series converges uniformly because the entries of $\mathbf B$ are Hölder continuous functions. (I agree with Yiorgos S. Smyrlis that they are actually real analytic, but I'm not going to prove that.) Here is a reference for the implication Hölder continuity $\implies$ uniform convergence; in the rest of the post I prove that $\mathbf B$ is Hölder continuous of order $1/2$.

Clearly, $\lambda(t)=\|\mathbf A(t)\|$ is Lipschitz continuous. It remains to show that $\mathbf x(t)$ is Hölder continuous. (The product of bounded $C^\alpha$ functions is $C^\alpha$).

Let $\delta(t)$ be the spectral gap of $\mathbf A(t)$, i.e. the difference between the largest and second largest eigenvalues. I claim that $\delta(t)$ is bounded from below by a positive constant. If not, then there is a bounded sequence $t_n$ along which $\delta(t_n)\to 0$. Select a convergent subsequence $t_n\to t_0$. The corresponding normalized eigenvectors $\mathbf x_n$ and $\mathbf{\tilde x_n} $ are orthogonal to each other. Passing to a further subsequence, we can make sure that $\mathbf x_n\to \mathbf x$ and $\mathbf{\tilde x_n} \to \mathbf{ \tilde x }$. It follows that $\mathbf x$ are $ \mathbf{ \tilde x }$ are both eigenvectors for $\lambda(t_0)$, and they are orthogonal, a contradiction.

Let $\delta>0$ be a uniform lower bound for the spectral gap. Fix $t$. Any unit vector $\mathbf u$ decomposes as $\mathbf u=\mathbf u'+\mathbf u''$ with $\mathbf u'=(\mathbf x(t)^H \mathbf u) \mathbf x(t)$ and $\mathbf u''\perp \mathbf x(t)$. The latter implies $|\mathbf A(t) \mathbf u''|\le (\lambda(t)-\delta)|\mathbf u''|$. Hence, $$ |\mathbf A(t)\mathbf u|^2 = |\mathbf A(t)\mathbf u'|^2 + |\mathbf A(t)\mathbf u''|^2 \le \lambda(t)^2 |\mathbf x(t)^H \mathbf u|^2+(\lambda(t)-\delta)^2 (1-|\mathbf x(t)^H \mathbf u|^2) $$ from where we conclude that $$ (1-|\mathbf x(t)^H \mathbf u|^2) \le \frac{\lambda(t)^2 - |\mathbf A(t)\mathbf u|^2}{2\delta \lambda(t)-\delta^2} \le \frac{\lambda(t)^2 - |\mathbf A(t)\mathbf u|^2}{\delta^2} \tag{1}$$ Apply (1) to $\mathbf u=\mathbf x(s)$, for which we have $$|\mathbf A(t)\mathbf x(s)|\ge \mathbf A(s)\mathbf x(s)-C|t-s| = \lambda(s)-C|t-s| \ge \lambda(t)-C|t-s|\tag{2}$$ (here and below $C$ is a generic constant, not necessarily the same at every occasion.) By virtue of (2), the right hand side of (1) is bounded by $C|t-s|$. Hence, $\mathbf x(t)^H \mathbf x(s) \ge 1-C|t-s|$, which implies
$$|\mathbf x(t)-\mathbf x(s)|^2 \le C|t-s|$$ as promised.

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