Let $f:[0,\infty) \mapsto [0, 1]$ be a non-increasing function and $W > 0$ be such that
$$ \int_{0}^\infty f(t)dt \le W. \tag{1} $$
Goal. I seek nontrivial upper-bounds on $f(t)$.
It is easy to deduce that
$$ f(t) \le \frac{W}{t},\forall t > 0. \tag{2} $$ Indeed, for every $t > 0$, we have $tf(t) \le \int_0^tf(s)ds \le \int_0^\infty f(s)ds \le W$.
Question. Apart from inequality (2), are there any nontrivial upper-bounds on $f(t)$ which can be obtained from (1) ?
For $t > W$, there is no better upper bound for $f(t)$. You can think of the following function \begin{equation*} f(x) = \begin{cases} W/t\qquad \text{if } x \in [0, t],\\ 0 \hspace{3.5em} \text{if } x > t. \end{cases} \end{equation*} It satisfies your condition with $f(t) = W/t$.
For $t < W$, the better upper bound is given by $1$ (since $W/t > 1$). In summary, the optimal upper bound for $f$ is $$f(t) \leq 1 \wedge W/t.$$
In addition, I think this optimal bound will not change even if you assume that $f$ is continuous.