Is this$$γ(h) = 1(h = 0) − 0.5 · 1(|h| = 2) − 0.25 · 1(|h| = 3)$$an autocovariance function?
How to check this? Is there a method one can use to check if a given function is an autocovariance funtion? I have found nothing in my "time series analysis" notes.
On
There is a another way to solve this.
We know that $\gamma$ is a valid autocovariance function iff, for every $n > 0$, the matrix $\Gamma_n$ defined as $\Gamma_n(i,j) = \gamma(|i-j|)$ is positive semidefinite.
In particular we must have that, for every $n > 0$, $\sum_{i,j} \Gamma_n(i,j) \geq 0$
Now let's assume $\gamma(h) = 0$ for all $h$ such that $|h|>3$ and $\gamma(0)=1$ (this includes your case).
Then for $n \geq 3$, $$ \sum_{i,j} \Gamma_n(i,j) = n (1 + 2 \gamma(1) + 2 \gamma(2) + 2 \gamma(3)) + C $$ where $C$ does not depend on $n$.
Since this has to be non-negative for every $n$, so must it be its limit for $n \to \infty$.
This implies that $$ \gamma(1) + \gamma(2) + \gamma(3) \geq - \frac 12 $$ which is not satisfied in your case (which is $\gamma(1) = 0, \gamma(2) = -0.5, \gamma(3) = -0.25$)
Nevermind I solved it.
For the spectral density we have $$f(\omega)=\frac{1}{2\pi}\sum_{h=-\infty}^{\infty}e^{-ih\omega}\rho(h)=\frac{1}{2\pi}(1-0.5e^{-i2\omega}-0.5e^{i2\omega}-0.25e^{-i3\omega}-0.25e^{i3\omega})=\frac{1}{2\pi}(1-\cos(2\omega)-0.5\cos(3\omega)),\mbox{ }\omega\in[-\pi,\pi].$$
For $\omega=0$ we have $f(0)=-\frac{1}{4\pi}<0$. The spectral density must be non-negative so this is not an ACVF.