Is there a quick way of determining whether the following functions are positive-definite or not?
$f: \mathbb{R}\to\mathbb{C}$
- $f(x) = 3$
- $f(x) = -3$
- $f(x) = x - 3$
- $f(x) = x + 3$
So, my attempts:
For $f(x) = x - 3$ if we take $n=1, a_1 = 1$ and $x_1$ arbitrary then $$1 \cdot f(x_1 - x_1) \cdot \overline{1} = 1 \cdot f(0) \cdot \overline{1} = 1 \cdot -3 \cdot \overline{1} = -3 < 0$$ so it is not.
Similar reasoning can be made on $f(x) = -3$. Is this part correct so far?
Now, $f(x) = 3$ seems to be positive definite but on the other hand if I take $a_1 = 1, a_2 = -1$ and arbitrary $x_1, x_2$ then $$a_1 f(x_1 - x_1) \overline{a_1} + a_1 f(x_1 - x_2) \overline{a_2} + a_2 f(x_2 - x_1) \overline{a_1} + a_2 f(x_2 - x_2) \overline{a_2} = \\ = 3 ( 1 \cdot \overline{1} + 1 \cdot \overline{-1} -1 \cdot \overline{1} -1 \cdot \overline{-1}) = 3 ( 1 - 1 - 1 + 1 ) = 0 $$ and it is not greater than zero! Does that mean the function is not positive definite?
What about $f(x) = x + 3$? Here I don't know how to proceed.
For real $x_i$'s we have $$\sum \sum a_ia_j ((x_i+3)-(x_j+3))$$ $$=\sum \sum a_ia_j (x_i-x_j)$$ $$= (\sum a_j) (\sum a_ix_i) -(\sum a_i) (\sum a_jx_j) =0$$ so $f$ is certainly not positive definite.