I am trying to solve the following exercise:

Determine the limit in $\mathcal{D}'(\mathbb{R})$ of $\lim_{t\rightarrow \infty} t^{2}xe^{itx}$, $u_{t} = t^{2}xe^{itx}$.

I have tried evaluating the integral $\int_{\mathbb{R}} t^{2}xe^{itx} \phi(x) dx $.

Making a change of variables I get:

$\int_{\mathbb{R}} y e^{iy} \phi(\frac{y}{t}) dy $,

this is convergent for each fixed $t$, however, as $t$ approaches infinity the support will be arbitrary large so the integral cannot be dominated easily by an integrable function in order to apply DCT.

I think one can rewrite the integral by using the fourier transform of $\delta$ as

$ \int_{\mathbb{R}} y e^{iy} \mathcal{F}(\delta(x))(y) \phi(\frac{y}{t}) dy $ = $ \int_{\mathbb{R}} \mathcal{F}(i\frac{d\delta(x-1)}{dx})(y) \phi(\frac{y}{t}) dy $

Can you then use that $<\mathcal{F} u , \phi> = <u, \mathcal{F}(\phi)>$ for distributions?

So that you would get something like

$ \int_{\mathbb{R}} (i\frac{d\delta(x-1)}{dx})(y) \mathcal{F}(\phi(\frac{y}{t}))dy $

and then use the definition of $\delta$ to get $<u_t , \phi> = -i\frac{d\mathcal{F}(\phi)}{dx}(\frac{1}{t})$

I am not sure about the details or if I have the right idea, any hints and help would be appreciated!


As pointed out in the comment, $t^2 xe^{itx} = -x\frac{\partial^{2}}{\partial x^{2}}e^{itx}$, using this I get:

$<u_t ,\phi> = \int_{\mathbb{R}} -x\frac{\partial^{2}}{\partial x^{2}}e^{itx} \phi(x) dx $, doing partial integration twice I get:

$\int_{\mathbb{R}} \frac{1}{t^2}e^{itx}(2\phi'(x) + x\phi''(x))dx = $

$ \frac{1}{t^2}(-i2\hat{\phi}(t) + (-i\frac{d}{dt}(-t^2\hat{\phi}(t))) =$

$ \frac{1}{t^2}(-i2\hat{\phi}(t) + i2t\hat{\phi}(t) + it^2\frac{d\hat{\phi}}{dt}) = $

$i \frac{d\hat{\phi}}{dt}(t)$

I don't really see what this would be when $t \rightarrow \infty$

  • 4
    $\begingroup$ $$t^2xe^{itx} = -x \frac{\partial^2}{\partial x^2} e^{itx}$$ $\endgroup$ – Daniel Fischer May 27 '14 at 12:55
  • $\begingroup$ I tried that but still I don't think I get it correctly. $\endgroup$ – El_Loco May 27 '14 at 17:45

For fixed $t\in\mathbb{R}$, and $\varphi\in \mathcal{D}(\mathbb{R})$, we have

$$\begin{align} \langle u_t,\varphi\rangle &= \int_\mathbb{R} t^2xe^{itx}\varphi(x)\,dx\\ &= - \int_\mathbb{R} \left(\frac{\partial^2}{\partial x^2} e^{itx}\right)\left(x\varphi(x)\right)\,dx \tag{integrate by parts twice}\\ &= - \int_\mathbb{R} e^{itx} \frac{\partial^2}{\partial x^2}\left(x\varphi(x)\right)\,dx. \end{align}$$

Now $\frac{\partial^2}{\partial x^2}\left(x\varphi(x)\right)\in \mathcal{D}(\mathbb{R})$, and the Riemann-Lebesgue lemma$^1$ says

$$\lim_{\lvert t\rvert\to\infty} \langle u_t,\varphi\rangle = 0.$$

$^1$ Overkill, for $C^1$ functions $f$ with compact support an integration by parts immediately shows that $\int_\mathbb{R} f(x) e^{itx}\,dx$ is $O(\lvert t\rvert^{-1})$.

  • $\begingroup$ Thank you, I totally forgot about Riemann-Lebesgue. $\endgroup$ – El_Loco May 27 '14 at 18:02

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