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Let $f$ be a complex valued Lebesgue integrable function over R, show that the function $g(t)= \int_{\mathbb R} f(x)\exp(-itx) \, dx$ is differentiable. I cannot seem to find a dominating $L^1$ function in order to apply differentiation under the integral sign. Any alternative ideas? Thanks

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    No, I require a dominating function for the derivative of the integrand, or for the difference f(x){exp(-itx)-exp(-isx)}/(t-s), I don't know how else to approach this problem.2012-12-13

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Fix an $x$. Consider $\frac{g(t+h) - g(t)}{h} = \int_{\mathbb{R}} f(x)e^{-itx}(\frac{e^{-ihx} - 1}{h}) dx$.

Notice that $\frac{e^{-ihx} -1}{h} = \frac{-i}{h} \int_0^{hx} e^{-is} ds$, from which we obtain the estimate $|\frac{e^{-ihx} - 1}{h}| \le 1$. So the integrand is bounded by $|f|$, and the differentiation under the integral sign follows by appealing to the dct.

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    You're correct. The estimate would actually be $|\$f$rac{e^{-ihx} - 1}{h}| \le |x|$. I apologize $f$or the hasty and incorrect response.2012-12-14
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There is something amiss with the question. For example, the functions $f_\alpha(x)=1/(x+i)^\alpha\,$ have Fourier transforms $\hat{f}_\alpha$ equal to $c_\alpha\cdot x^{\alpha-1}e^{-x}$ for $x>0$, and $0$ for $x<0$, for constants $c_\alpha$, for $\Re(\alpha)>1$ at least. (This is most easily computed by computing the Fourier transform of the function that is $0$ for $x<0$ and $x^{\alpha-1}e^{-x}$ for $x>0$, by Gamma-function machinations.) For $\alpha=1+\epsilon\,$ the function is in $L^1\cap L^2$, but its Fourier transform is not differentiable at $0$. (That is, the conclusion of differentiability is not correct as posed...)