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Let $f(x)$ be a smooth function from $\mathbb{R}$ to $\mathbb{R}$. Let $\mu$ be a compactly supported Borel measure (not necessarily positive) on $\mathbb{R}^n$. Define $$ \tilde{\mu}(\xi) = \int e^{i (\xi_1 f(x_1) + \ldots + \xi_n f(x_n))} \mu(dx). $$ I have estimates of the form $|\tilde{\mu}(\xi)| \leq C(\xi)$ for any $\xi \in \mathbb{C}^n$. My question is if it possible to find the similar estimates on Fourier transform $$ \hat{\mu}(\xi) = \int e^{i \xi x} \mu(dx) $$ using given estimates on $\tilde{\mu}(\xi)$?

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Yes, thank you, it was an errata. –  Nimza Oct 18 '12 at 10:19

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As stated, no. Let $f \equiv 0$, and $\mu$ with $\int \mu = 0$. You then have $|\tilde{\mu}(\xi)| = 0$ for any $\xi$, and you have no control over the non-zero frequencies $\hat{\mu}$.

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Of course it is'n true in general. There may be some conditions on $f$ and $\mu$ which garantee the possibility of estimation of Fourier transform. I don't know about such conditions and I would like to know. For example, for function $f(x) = x^{3}$. –  Nimza Oct 18 '12 at 10:19

If your $f$ is one-to-one, make the change of the variable in your integral $y_i=f(x_i)$. You will obtain a new measure $\mu_1$, also with compact support, and your $\tilde\mu$ will be the ordinary Fourier transform of this new measure $\mu_1$.

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Yes, the task is exactly to find estimates on $\mu$ Fourier transform if we have estimates on $\mu_1 = \phi_{*}\mu$ Fourier transform, where $phi(x_1,\ldots,x_n) = (f(x_1),\ldots,f(x_n))$ and $\phi_{*}\mu$ is a pushforward of measure $\mu$. –  Nimza Oct 18 '12 at 19:34
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I am not sure what kind of estimates you are talking about. If you mean the rate of decrease at infty, this is related to the smoothness of your measure. Then the answer depends on the smoothness of $f$. can you be more specific? –  Alexandre Eremenko Oct 18 '12 at 23:10
    
I'm talking about similar estimates of Fourier transform $\hat{\mu}(\xi)$. I want to find such function $K(\xi)$ that $|\hat{\mu}(\xi)| \leq K(\xi)$. I don't want any estimates on infinity so methods like stationary phase e.t.c. can't help. –  Nimza Oct 20 '12 at 9:10

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