Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Long story short, the question I'm stuck on is as follows:

Let $f$ be a positive-definite function. Prove that if $f$ is continuous at $0$, then it is continuous everywhere.

Here's the long version:

We say that a function $f:\mathbb{R}\to \mathbb{C}$ is positive definite if the matrix $A_f[\{t_1,t_2,\dots,t_n\}]$, whose entries are given by $$ A_f[\{t_1,t_2,\dots,t_n\}]=[f(t_i-t_j)]_{i,j=1}^n $$ Is positive semidefinite for all choices of $t_1,\dots,t_n \in \mathbb{R}$. In the whole problem, we are meant to show that $f$ has the following properties:

  • $f(-t) = \overline{f(t)}$
  • $f(0) \in \mathbb{R}$ and $f(0) \geq 0$
  • $|f(t)|\leq f(0)$ for all $t \in \mathbb{R}$
  • if $f$ is continuous at $0$, then it is continuous everywhere

The first three parts may all be solved by considering the $2\times 2$ matrix $A_f[0,t]$ where $t\in \mathbb{R}$ is arbitrary. Because $A_f[0,t]$ is Hermitian, the first statement holds. Because $A_f[0,t]$ must have non-negative trace, we conclude that the second statement holds. Becuase $A_f[0,t]$ has a non-negative determinant, we conclude that the third statement holds. That fourth statement, however, has me stumped.

As far as I can tell, there is no more insight to be gleaned from $2\times 2$ matrices. Presumably, I need to find an upper bound for $|f(t) - f(t+\delta)|$ given that $|f(\delta) - f(0)|$ can be made arbitrarily small. I've noticed that $\det A_f[0,t,t+\delta]$ can be finagled into something like $f(0)|f(t) - f(t+\delta)|^2$. However, it's not clear to me how I would use this to the desired ends.

There's also a good chance that I've managed to think myself into a hole, given that this one small part of one problem has given me more trouble than the rest of the assignment. The question claims that this problem can be solved using the fact that a semi-definite matrix has a non-negative trace and determinant, and that all principal submatrices have a non-negative determinant.

I think that just about covers it. If you've made it this far, thank you for your time; I tried not to make this a wall of text. Any helpful nudges in the right direction would be very much appreciated; an attempt at an answer doubly so.

share|improve this question
1  
Having finally typed out this question, I wondered if there was a hint for this problem in the back of the book. There was. It turns out I'm on the right track now that I am (finally) considering $3\times 3$ matrices (the hint is simply "consider the $n=3$ case."). Things are looking much more hopeful now, but I don't feel like the solution is quite at hand. At any rate, I plan to sleep on it. All input is, however, still (and always) welcome. –  Omnomnomnom Nov 17 '13 at 6:38

1 Answer 1

up vote 3 down vote accepted

Hint: The matrix $$ A=\pmatrix{ f(0) &f(-t) &f(-t-h)\\ f(t) &f(0) &f(-h)\\ f(t+h) &f(h) &f(0)} $$ is congruent to $$ B=\pmatrix{ f(0) &f(-t) &f(-t-h)-f(-t)\\ f(t) &f(0) &f(-h)-f(0)\\ f(t+h)-f(t) &f(h)-f(0) &2f(0)-f(h)-f(-h)}. $$ Now consider the $2\times2$ submatrix taken from the entries at the four corners of $B$.

share|improve this answer
1  
Brilliant idea to use a congruent matrix! I know it's an elegant solution because I can't help but feeling stupid for not having thought of it myself (I had considered matrix similarity, which can be used to the same effect, but had only considered it in the $2\times 2$ case, and not quite with this end in mind). This will definitely go into my bag of tricks. Thank you. –  Omnomnomnom Nov 17 '13 at 14:56

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.