$\int_{\mathbb{R}}f(x)e^{-ixz}d\mu_x$ analytic for $f\in L_1$

Question

Let $f\in L_1(-\infty,\infty)$ be a Lebesgue-summable function on $\mathbb{R}$ and let $x\mapsto e^{\delta|x|}f(x)$ also be Lebesgue-summable on all the real line. I have added the condition that $f\in L_1(-\infty,\infty)$, but I think that if $e^{\delta|x|}f(x)$ is Lebesgue-summable (the only condition explicitly stated in the book, Kolmogorov-Fomin's, p. 430 here) on $\mathbb{R}$ then $f$ also is.

Kolmogorov-Fomin's Элементы теории функций и функционального анализа states (p. 430) that the complex function defined by $$g(z)=\int_{\mathbb{R}}f(x)e^{-ixz}d\mu_x$$where the integral is Lebesgue's, $\mu_x$ is Lebesgue linear measure and $z=\lambda+i\nu$, $\lambda,\nu\in\mathbb{R}$ is a complex variable, is analytic in $\{z\in\mathbb{C}:|\nu|<\delta\}$. I clearly see that the integral is the Fourier transform of $f(x)e^{\nu x}$.

By trying to prove the analiticity of $g$ I have convinced myself that, if $x\mapsto e^{\delta|x|}f(x)$ is Lebesgue-summable on $\mathbb{R}$, then $x\mapsto xf(x)$ also is. If that is correct, then I think we could use the Cauchy-Riemann equations and calculate the two partial derivatives with respect to $\lambda$ by using$^1$ the fact, where I call $F$ the Fourier transform, that $\frac{d}{d\lambda}F[e^{\nu x}f(x)](\lambda)=-iF[xe^{\nu x}f(x)]$. Nevertheless, I have no idea about how to calculate $\frac{\partial}{\partial\nu}\text{Re }g$ and $\frac{\partial}{\partial\nu}\text{Im }g$ (I do not even see why such derivatives exist).

Can anybody prove the analiticity of $g$ in $\{z\in\mathbb{C}:|\nu|<\delta\}$ either by using the Cauchy-Riemann equations or in some other way? I $\infty$-ly thank you!

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$^1$To complete "what I have tried" I should say that I get the two following partial derivatives:$$\frac{\partial}{\partial\lambda}\text{Re }g(\lambda)=\int_{\mathbb{R}}xe^{\nu x}[\text{Im }f(x)\cos(x\lambda)-\text{Re }f(x)\sin(x\lambda)]d\mu_x$$ $$\frac{\partial}{\partial\lambda}\text{Im }g(\lambda)=\int_{\mathbb{R}}xe^{\nu x}[\text{Re }f(x)\cos(x\lambda)+\text{Im }f(x)\sin(x\lambda)]d\mu_x$$

Answer 1

Yes, the fact that $h(x) = e^{\delta\lvert x\rvert}f(x)$ is Lebesgue integrable implies that $f$ is also Lebesgue integrable. Since $x\mapsto e^{\delta \lvert x\rvert}$ is continuous and everywhere $\geqslant 1$, $f$ is (Lebesgue) measurable if and only if $h$ is measurable, and since $\lvert f(x)\rvert \leqslant \lvert h(x)\rvert$, the finiteness of $\int \lvert f\rvert\,d\mu$ follows from $\int \lvert h\rvert\,d\mu < +\infty$.

One nice way to prove $g$ analytic uses Morera's theorem. First we note that $\lvert e^{-ixz}\rvert = e^{\operatorname{Re} (-ixz)} = e^{x\cdot \operatorname{Im} z} < e^{\delta\lvert x\rvert}$, so the integrand in the definition of $g(z)$ is dominated by $\lvert h\rvert$ for all eligible $z$. The integrand also depends continuously on $z$, so by the dominated convergence theorem, $g$ is continuous. Now we use Morera's theorem to conclude that $g$ is analytic: Let $\Delta$ be a closed triangle contained in the strip $S = \{ z : \lvert \operatorname{Im} z\rvert < \delta\}$. Then

\begin{align} \int_{\partial \Delta} g(z)\,dz &= \int_{\partial\Delta} \int_\mathbb{R} f(x) e^{-ixz}\,d\mu_x\,dz\\ &= \int_\mathbb{R} f(x) \int_{\partial\Delta} e^{-ixz}\,dz\,d\mu_x \tag{Fubini}\\ &=\int_\mathbb{R} f(x)\cdot 0\,d\mu_x \tag{Goursat}\\ &= 0, \end{align}

so Morera's theorem says that $g$ is analytic in $S$.

Another way uses differentiation under the integral sign. For that, we must see that for every $0 < \delta_0 < \delta$, there is a constant $C < +\infty$ such that

$$\lvert x\cdot e^{-ixz}\rvert \leqslant C\cdot e^{\delta\lvert x\rvert}$$

for all $x\in\mathbb{R}$ and all $z$ with $\lvert \operatorname{Im} z\rvert \leqslant \delta_0$. That follows from the observation that $\lvert x\rvert\cdot e^{-\alpha\lvert x\rvert}$ is bounded for every $\alpha > 0$, where we use $\alpha = \delta - \delta_0$. Then the dominated convergence theorem (or a variant or corollary thereof, depends on how the text is organised) allows differentiation under the integral. One can now verify the Cauchy-Riemann equations in the real form, but it is less cumbersome to check them in the complex form,

$$\frac{\partial g}{\partial \overline{z}} \equiv 0,$$

since we can also move the $\frac{\partial}{\partial\overline{z}}$ operator under the integral sign, and obtain

\begin{align} \frac{\partial g}{\partial\overline{z}}(z) &= \frac{\partial}{\partial\overline{z}}\int_{\mathbb{R}} f(x) e^{-ixz}\,d\mu_x\\ &= \int_{\mathbb{R}} \frac{\partial}{\partial\overline{z}}\left(f(x) e^{-ixz}\right)\,d\mu_x\\ &= \int_{\mathbb{R}} f(x) \frac{\partial}{\partial\overline{z}} e^{-ixz}\,d\mu_x\\ &= \int_\mathbb{R} f(x)\cdot 0\,d\mu_x\\ &= 0, \end{align}

since $z\mapsto e^{-ixz}$ is holomorphic for every fixed $x$.

Answer 2

You can directly compute the derivative of $g$ by dominated convergece

$\lim_{h\to 0}\dfrac{g(z+h) - g(z)}{h}=\lim_{h\to 0}\int_{\mathbb{R}}f(x)\dfrac{e^{-ix(z+h)}- e^{-ixz}}{h}d\mu_x$

And we have

$$\dfrac{e^{-ix(z+h)}- e^{-ixz}}{h} = e^{-ixz} \dfrac{e^{-ixh}- 1}{h}$$

Let $z = \lambda + i\nu$ with $|\nu| < \delta$

$$\left|e^{-ixz} \dfrac{e^{-ixh}- 1}{h}\right| \leq e^{|\nu| |x|} \left|\dfrac{e^{-ixh}- 1}{h}\right| \leq e^{|\nu| |x|} \left|\dfrac{e^{|x||h|}- 1}{h}\right| \leq e^{|\nu| |x|} \dfrac{e^{(\delta - |\nu|)|x|}-1}{\delta-|\nu|}, \forall |h| < \delta -|\nu|$$

i.e.

$$\left|e^{-ixz} \dfrac{e^{-ixh}- 1}{h}\right| \leq \dfrac{e^{\delta|x|}-e^{|\nu||x|}}{\delta-|\nu|}, \forall |h| < \delta -|\nu|$$

Then $$\left|f(x)\dfrac{e^{-ix(z+h)}- e^{-ixz}}{h}\right| \leq |f(x)|\dfrac{e^{\delta|x|}-e^{|\nu||x|}}{\delta-|\nu|} \leq |f(x)|\dfrac{e^{\delta|x|}}{\delta-|\nu|}, \forall |h| < \delta -|\nu|$$

The last term at RHS is integrable by your assumption, so you can apply dominated convergence to conclude