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I am reading a proof that states that the Fourier transform $\hat{f}$ of a function $f\in L^1$ is continuous and bounded (here $f:\mathbb{R}^d\to \mathbb{R}$). I am having trouble with understanding the use of dominated convergence theorem in the continuous part.

Here is the proof, as appears in the notes: Let $(\zeta_n)$ be a sequence in $\mathbb{R}^d$ such that $\zeta_n\to\zeta\in\mathbb{R}^d$. Then $(\exp(i\langle x,\zeta\rangle)−\exp(i\langle x,\zeta_n\rangle)f(x)\to 0$ as $n\to\infty$ for each $x\in\mathbb{R}^d$. Since $|\hat{f}(\zeta)− \hat{f}(\zeta_n)|\leq 2\|f\|_1$, an application of the Dominated convergence theorem shows that $\hat{f}(\zeta_n)\to\hat{f}(\zeta)$.

The trouble I am having is with the use of the DCT. Namely, isn't the goal to show $\exp(\langle x,\zeta_n\rangle)f(x)$ is dominated by some integrable function? How does $|\hat{f}(\zeta)− \hat{f}(\zeta_n)|\leq 2\|f\|_1$ enable us to use the DCT?

Thank you in advance. B.

Edit: It appears $|\exp(\langle x,\zeta_n\rangle)f(x)|\leq |f(x)|$ which is integrable, so shouldn't this suffice?

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  • $\begingroup$ You need to bound the integrand not the $|\hat{f}(\zeta)-\hat{f}(\zeta_n)|$ so your suspicion about that argument not working is not unfounded. Your edit is correct. Also you should use \langle and \rangle for $\langle$ and $\rangle$, respectively, when writing inner products. It is much cleaner. $\endgroup$ Commented Sep 8, 2015 at 16:37

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So you take your sequence $\zeta_n$ in $\mathbb{R}^d$ converging to some $\zeta\in\mathbb{R}^d$. Then $$|\hat{f}(\zeta_n)-\hat{f}(\zeta)|\leq \int_{\mathbb{R}^d}|f(x)|\left|\exp(i\langle x,\zeta_n\rangle) - \exp(i\langle x,\zeta\rangle)\right|\,dx$$ Now the sequence of integrands converges to zero for each $x$, and it is bounded by $2|f(x)|$ for every $x$. As $f\in L^1(\mathbb{R}^d)$, the DCT then asserts that the sequence of integrals also tends to zero.

In fact, a stronger assertion is true, with almost the same proof: The fourier transform of an $L^1$ function is uniformly continuous in $\mathbb{R^d}$.

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  • $\begingroup$ Is it possible to say something if we take a net instead of a sequence? $\endgroup$ Commented Nov 14, 2019 at 11:10

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