Let $S$ is finite, $S \subset \mathbb{Z}$ and $f(n)=g(n)=0,\, n \in \mathbb{Z} \setminus S$.
We use a standart notation $\|f\|_{p} = \left( \sum_{i\in S} |f(i)|^p \right)^{\frac{1}{p}}, p < \infty;\, \|f\|_{\infty} = \max_{i\in S}|f(i)| $. We have:
(1) $ \sum_{n\in\mathbb{Z}}h(n) \geq \|f\|_{\infty} \|g\|_1 $.
Why? If $\|f\|_{\infty} = f(n_0) \, \Rightarrow \, h(n_0 + n) \geq \|f\|_{\infty} g(n)$ and summing up we get (1).
In the same way:
(2) $ \sum_{n\in\mathbb{Z}}h(n) \geq \|g\|_{\infty} \|f\|_1 $.
From (1) and (2) follows:
(3) $ \sum_{n\in\mathbb{Z}}h(n) \geq \frac{1}{q} \|f\|_{\infty} \|g\|_1 + \frac{1}{p} \|g\|_{\infty} \|f\|_1 $.
Now we write:
$\sum_{n\in S} f(n)^p \leq \|f\|_{\infty}^{p-1}\|f\|_1$, which implies:
(4) $ \|f\|_{\infty}^{\frac{1}{q}} \|f\|_1^{\frac{1}{p}} \geq \|f\|_p $.
(5) $\|g\|_{\infty}^{\frac{1}{p}} \|g\|_1^{\frac{1}{q}} \geq \|g\|_q $.
Multiplying (4) and (5) we get:
(6) $ \left( \|f\|_{\infty} \|g\|_1 \right)^{\frac{1}{q}} \left( \|g\|_{\infty} \|f\|_1 \right)^{\frac{1}{p}} \geq \|f\|_p \|g\|_q $.
Now, (3) , (6) and weighted AM-GM imply:
$ \sum_{n\in\mathbb{Z}}h(n) \geq \|f\|_p \|g\|_q $.
Comment. The fact that $f,g$ vanish with the exception of finitely many integers is not so essential. Slightly restating the problem we can get rid of that requirement.