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

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!

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

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

• Thank you so much very clear and detailed answer! There's just one thing I don't grasp: how can we see that $|e^{-ixh}-1|\leq|e^{|xh|-1}|$? Thank you again!!! – Self-teaching worker Nov 21 '14 at 21:01
• @DavideZena $|e^{-ixh} -1| = |\sum_{n=1}^\infty \dfrac{(-ixh)^n}{n!}| \leq \sum_{n=1}^\infty|\dfrac{(-ixh)^n}{n!}| = \sum_{n=1}^\infty\dfrac{|xh|^n}{n!} = e^{|xh|} - 1$ – Petite Etincelle Nov 21 '14 at 21:05
• Very, very clear: I heartily thank you! – Self-teaching worker Nov 21 '14 at 21:21
• @DavideZena You are welcome :) – Petite Etincelle Nov 21 '14 at 21:23
• @DavideZena Yeah, dominated convergence is a very useful tool – Petite Etincelle Nov 21 '14 at 22:19

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 $x$.

• Complex line integrals are also/can also be interpreted as Lebesgue integrals. You can pull it back to an integration over a real interval via a parametrisation of the contour, $\int_\gamma h(z)\,dz = \int_0^1 h(\gamma(t))\cdot \gamma'(t)\,dt$, or you can view it as a Lebesgue integral with respect to the complex measure $dz$. For the application of Fubini's theorem, you need the measurability of the integrand, and the finiteness of $$\int_{\partial\Delta}\int_\mathbb{R} \lvert f(x)e^{-ixz}\rvert\,d\mu_x\,\lvert dz\rvert,$$ but since $\int_{\mathbb{R}}\lvert f(x)e^{-ixz}\rvert\,d\mu_x$ is ... – Daniel Fischer Nov 21 '14 at 21:57
• ... bounded independently of $z\in S$ since the integrand is dominated by $\lvert h(x)\rvert$, the finiteness of the length of $\partial\Delta$ yields that. Since the integrand depends continuously on $z$, we have pointwise convergence of the integrands as $z_k\to z$, and since we have the global domination by $\lvert h(x)\rvert$, the dominated convergence theorem asserts the continuity of the integral. For the differentiation under the integral sign, we need that the difference quotients are dominated (by a constant multiple of $\lvert h(x)\rvert$), then the DCT shows that the difference ... – Daniel Fischer Nov 21 '14 at 21:58
• ... quotient of the integrals, which is the integral of the difference quotients by linearity, converges to the integral of the partial derivative. Sometimes the conditions for differentiation under the integral sign are explicitly given as a corollary or a second theorem, or even as examples in the statement of the main theorem, since it's an important application of that theorem. – Daniel Fischer Nov 21 '14 at 21:58
• @DavideZena Oops, forgot to ping you. – Daniel Fischer Nov 21 '14 at 22:07
• Yes, it's correct. Usually, one cares about the function being differentiable not only at $y_0$ but on a full neighbourhood of $y_0$, and then, rather than dominating the difference quotients, one posits that $$\left\lvert \frac{\partial f}{\partial y}(x,y)\right\rvert \leqslant \varphi(x)$$ for [almost] all $x$, for all $y\in (y_0-\delta,y_0+\delta)$, which is typically simpler to express and check. But if only differentiability of $f(x,\,\cdot\,)$ in $y_0$ is given, that of course isn't possible. – Daniel Fischer Nov 22 '14 at 13:25