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By definition, $\tilde{f}(0)$ exists if and only if $\int_0^\infty f(x)/x\,dx$ exists. As $f(x)$ is smooth and supported on $[0,2]$, we have that that $$|f(x)|=\left|\int_0^x f'(t)\,dt\right|\leq x\sup_{[0,x]}|f'|,\qquad x\geq 0.$$ Hence $f(x)/x$ is bounded, so $\tilde{f}(0)$ exists (and can be bounded easily in terms of $f'$). In fact, since all the derivatives of $f$ vanish at $0$, Taylor's theorem (with explicit formula for the remainder term) shows that $$|f(x)|\leq\frac{x^k}{k!}\sup_{[0,x]}|f^{(k)}|,\qquad x\geq 0,\qquad k\in\mathbb{Z}_{\geq 0}.$$ So $f(x)$ decays very rapidly as $x\to 0+$, and (in the same way) also as $x\to 2-$.