Return to Answer

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 $$|f(x)|=\left|\int_0^x f'(t)\,dt\right|\leq x\sup_{0\leq t\leq x}|f'(t)|,\qquad x\geq 0.$$ Hence $f(x)/x$ is bounded, so $\tilde{f}(0)$ exists (and can be bounded easily in terms of $f'$).

More generally, since all the derivatives of $f$ vanish at $0$, repeated integration by parts shows that (cf. Taylor's theorem) $$|f(x)|=\left|\int_0^x \frac{(x-t)^{k-1}}{(k-1)!}f^{(k)}(t)\,dt\right|\leq\frac{x^k}{k!}\sup_{0\leq t\leq x}|f^{(k)}(t)|,\quad x\geq 0,\quad k\in\mathbb{Z}_{\geq 1}.$$ So $f(x)$ decays very rapidly as $x\to 0+$, and (in the same way) also as $x\to 2-$. This shows, incidentally, that $\tilde f(w)$ converges absolutely for every $w\in\mathbb{C}$ (and it decays rapidly in every vertical strip).

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 $$|f(x)|=\left|\int_0^x f'(t)\,dt\right|\leq x\sup_{0\leq t\leq x}|f'(t)|,\qquad x\geq 0.$$ Hence $f(x)/x$ is bounded, so $\tilde{f}(0)$ exists (and can be bounded easily in terms of $f'$).

More generally, since all the derivatives of $f$ vanish at $0$, repeated integration by parts shows that (cf. Taylor's theorem) $$|f(x)|=\left|\int_0^x \frac{(x-t)^{k-1}}{(k-1)!}f^{(k)}(t)\,dt\right|\leq\frac{x^k}{k!}\sup_{0\leq t\leq x}|f^{(k)}(t)|,\quad x\geq 0,\quad k\in\mathbb{Z}_{\geq 1}.$$ So $f(x)$ decays very rapidly as $x\to 0+$, and (in the same way) also as $x\to 2-$. Of course the same is true of every derivative of $f(x)$. Hence in fact $\tilde f(w)$ converges absolutely for every $w\in\mathbb{C}$, and it decays rapidly in every vertical strip.

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 $$|f(x)|=\left|\int_0^x f'(t)\,dt\right|\leq x\sup_{0\leq t\leq x}|f'(t)|,\qquad x\geq 0.$$ Hence $f(x)/x$ is bounded, so $\tilde{f}(0)$ exists (and can be bounded easily in terms of $f'$). More generally, since all the derivatives of $f$ vanish at $0$, repeated integration by parts shows that (cf. Taylor's theorem) $$|f(x)|=\left|\int_0^x \frac{(x-t)^{k-1}}{(k-1)!}f^{(k)}(t)\,dt\right|\leq\frac{x^k}{k!}\sup_{0\leq t\leq x}|f^{(k)}(t)|,\quad x\geq 0,\quad k\in\mathbb{Z}_{\geq 1}.$$ So $f(x)$ decays very rapidly as $x\to 0+$, and (in the same way) also as $x\to 2-$.

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 $$|f(x)|=\left|\int_0^x f'(t)\,dt\right|\leq x\sup_{0\leq t\leq x}|f'(t)|,\qquad x\geq 0.$$ Hence $f(x)/x$ is bounded, so $\tilde{f}(0)$ exists (and can be bounded easily in terms of $f'$). 

More generally, since all the derivatives of $f$ vanish at $0$, repeated integration by parts shows that (cf. Taylor's theorem) $$|f(x)|=\left|\int_0^x \frac{(x-t)^{k-1}}{(k-1)!}f^{(k)}(t)\,dt\right|\leq\frac{x^k}{k!}\sup_{0\leq t\leq x}|f^{(k)}(t)|,\quad x\geq 0,\quad k\in\mathbb{Z}_{\geq 1}.$$ So $f(x)$ decays very rapidly as $x\to 0+$, and (in the same way) also as $x\to 2-$. This shows, incidentally, that $\tilde f(w)$ converges absolutely for every $w\in\mathbb{C}$ (and it decays rapidly in every vertical strip).

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 $x\to 2-$.

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 $$|f(x)|=\left|\int_0^x f'(t)\,dt\right|\leq x\sup_{0\leq t\leq x}|f'(t)|,\qquad x\geq 0.$$ Hence $f(x)/x$ is bounded, so $\tilde{f}(0)$ exists (and can be bounded easily in terms of $f'$). More generally, since all the derivatives of $f$ vanish at $0$, repeated integration by parts shows that (cf. Taylor's theorem) $$|f(x)|=\left|\int_0^x \frac{(x-t)^{k-1}}{(k-1)!}f^{(k)}(t)\,dt\right|\leq\frac{x^k}{k!}\sup_{0\leq t\leq x}|f^{(k)}(t)|,\quad x\geq 0,\quad k\in\mathbb{Z}_{\geq 1}.$$ So $f(x)$ decays very rapidly as $x\to 0+$, and (in the same way) also as $x\to 2-$.