TITLE: Prove that if positive-definite $f$ is continuous at $0$, it is continuous on $\mathbb{R}$
QUESTION [13 upvotes]: 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.

REPLY [6 votes]: 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 $B$.