Proving $\int^{t}_{-\infty}f(x)e^{-2\pi iz(x-t)}dx$ is bounded in $\Bbb{C}$ (i.e. respect to $z$)

Question

Suppose $f$ is continuous and of moderate decrease, i.e. $\exists A\; s.t.|f(x)|\leq \frac{A}{1+x^2} \; \forall x \in \Bbb{R}$ and $\hat f(\zeta)=0 ,\;\forall \zeta \in \Bbb{R}$. For each fixed real number $t$, let $A(z) = \int^{t}_{-\infty}f(x)e^{-2\pi iz(x-t)}dx$ and $B(z) = -\int^{\infty}_{t}f(x)e^{-2\pi iz(x-t)}dx$, define $F$ in $\Bbb{C}$ s.t. $F=A$ in the closed upper half plane and $F=B$ in the lower half plane. Prove that $F$ is entire and bounded.

I know how to prove $F$ is entire, by simply apply symmetry principle. But I have trouble in proving that $F$ is bounded. I think for $Im(z)<0$ I can use $F(z)=B(z)$ to estimate $F(z)$ when $|z|\to \infty$ and use $F(z)=A(z)$ for $Im(z)>0$, but I simply lost in the middle. Could you please help me? Thank you.

Answer 1

If boundedness is the only thing bothering you, then notice that for $\Im z \ge 0$ (and $z = a + b \Bbb i$) we have, for every $t \in \Bbb R$,

$$|F(z)| = |A(z)| = \left| \int \limits _{-\infty} ^t f(x) \ \Bbb e ^{2 \pi \Bbb i z (x-t)} \Bbb d x \right| \le \int \limits _{-\infty} ^t \left| f(x) \ \Bbb e ^{2 \pi \Bbb i z (x-t)} \right| \Bbb d x = \int \limits _{-\infty} ^t |f(x)| \ \Bbb e ^{2 \pi b (x-t)} \ \Bbb d x \le \int \limits _{-\infty} ^t \frac A {1 + x^2} \cdot 1 \ \Bbb d x = A \left( \arctan t + \frac \pi 2 \right) .$$

(We have $\Bbb e ^{2 \pi b (x-t)} \le 1$ because $b \ge 0$ and $x \le t$.)

Similarly, for $\Im z \le 0$ you will get

$$|F(z)| \le A \left( \frac \pi 2 - \arctan t \right) ,$$

whence it follows that $|F|$ is bounded by $A \max \left\{ \arctan t + \frac \pi 2, \frac \pi 2 - \arctan t \right\}$, which no longer depends on $z$.

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There is something interesting, though: since $f$ may be viewed as a tempered distribution, and since $\hat f = 0$, it follows that $f = 0$, hence $F=0$.

Answer 2

We have $$ \left\lvert \int^t_{-\infty} f(x) e^{-2\pi i z(x-t)} dx \right \rvert \le \int^t_{-\infty}\lvert f(x) \rvert \left \lvert e^{-2\pi i z(x-t)} \right \rvert dx \le A \int^t_{-\infty} \frac{\left \lvert e^{-2\pi i z(x-t)}\right \rvert}{1+x^2} dx$$ Now $\left \lvert e^{-2\pi i z(x-t)} \right \rvert = e^{2\pi b(x-t)}$ where $b= \text{Im}\{z\} \ge 0$ since this definition applies for $z$ in the upper half plane. Then $$\left\lvert \int^t_{-\infty} f(x) e^{-2\pi i z(x-t)} dx \right \rvert \le Ae^{-2\pi bt}\int^t_{-\infty } \frac{e^{2\pi bx}}{1+x^2}dx.$$ The right hand side is a continuous function of $b$, hence bounded on any compact set: $b \in [0,R], \,\, R > 0$ . Thus it suffices to bound the expression for large $b$. For $b$ away from zero, we see $$\int^t_{-\infty } \frac{e^{2\pi bx}}{1+x^2}dx \le \int^t_{-\infty} e^{2\pi b x} dx = \left.\frac{1}{2\pi b}e^{2\pi b x}\right|_{x=-\infty}^{x=t} = \frac{e^{2\pi bt}}{2\pi b}.$$ Plugging this in above gives $$\left\lvert \int^t_{-\infty} f(x) e^{-2\pi i z(x-t)} dx \right \rvert \le \frac{A}{2\pi b} \le \frac{A}{2\pi} $$ for $b\ge 1$. Combine this with whatever bound you get for $b \in [0,1]$ to see that $A(z)$ is bounded. A nearly identical manipulation shows that $B(z)$ is bounded for $\text{Im}\{z\} \le 0$.