Maximum of a positive definitive function

Question

Let $f$ be a positive definitive function from $ \mathbb{R}^d \to \mathbb{R}$.

in https://en.wikipedia.org/wiki/Positive-definite_function it is stated that $ |f(x)| \leq f(0) $.

I couldnt find any proof. Can you help me?

Answer 1

Im using the same notation as in the wikipedia article you linked.

The first point $f(0) \geq 0$ follows by taking $n=1$ in the definition and an arbitrary $x \in \mathbb{R}$: Then the "Matrix" $f(0)$ has to be positive semi-definite, which means that $f(0) \geq 0$.

Now the second point: Here we take $n=2$ in the definition and choose $x_1=x$ and $x_2 = 0$. Now you get $$a_{11} = f(0), \ a_{12} = f(x), \ a_{21} = f(-x) = \overline{f(x)}, \ a_{22} = 0$$ and therefore the matrix $$ \left[ {\begin{array}{cc} f(0) & f(x) \\ \overline{f(x)} & f(0) \\ \end{array} } \right]$$

has to be positive semi-definite which in particular means that the determinant has to be positive. So we get $f(0)² - |f(x)|² \geq 0$ and therefore $f(0) \geq |f(x)|$ because of the first point.