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Assume that $f(x) \leq \frac A {1+x^2}$ on $\mathbb R$. I now want to show that $$ \hat f(\xi) = \int_{\mathbb R} f(x) \exp (-2 \pi i \xi x) dx $$ is continuous. By some simple calculations I get $$ |\hat f(\xi) - \hat f(\xi+h)| \leq A \int_{-\infty}^\infty \frac {|1-e^{-2\pi ixh}|}{1+x^2} dx \leq A \int_{-\infty}^\infty \frac {|1-\cos(2\pi x h)|}{1+x^2} dx $$ I wanted to use the Dominated convergence theorem. How can I apply this there ? Thanks in advance.

  • You want to show that the last integral vanishes as $h \rightarrow 0$. What are the hypotheses of the dominated convergence theorem? Does this integral satisfy those hypotheses? – dls Jun 25 '13 at 17:10
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    If you have the Dominated convergence theorem, isn't it clear? Fix $\xi$. Let $g_h(x) = f(x)\exp(-2\pi (\xi+h)x)$. The $g_h$ converge pointwise to $g_0$, and are dominated by $\frac{A}{1+x^2}$. – Daniel Fischer Jun 25 '13 at 17:10

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