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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. In fact repeating the above argument for $f'$, $f''$, etc. in place of $f$, we see that $$|f(x)|\leq x^n\sup_{[0,x]}|f^{(n)}|,\qquad x\geq 0.$$ So $f(x)$ decays very rapidly as $x\to 0+$, and (in the same way) also as $x\to 2-$.