Is this operator a Fourier multiplier operator?

Question

I want to study the Fourier transform of $$L_{\alpha}(t) = \frac{e^{i\alpha t}}{t^2} - i\frac{\alpha}{t}$$

Basically i am trying to get a grip on, given a $f$, what is $f(t)\ast L_{\alpha}(t)$ and am looking at $L_{\alpha}\ast$ as a Fourier multiplier operator.

Motivation : Here

My main motivation was taking derivative of $f\ast\text{p.v.}\frac{e^{i\alpha t}}{t}$ and from comments I realize that its more appropriate to take a distributional derivative of this itself. So my question is on derivative of $f\ast\text{p.v.}\frac{e^{i\alpha t}}{t}$

Answer 1

Ok, from a comment above, I assume that this is what you should do in order to have your operator as a Fourier Multiplier one (or, more essentially, as a well-defined one!)

STEP 1: First, you should correct a bit the notion of $f * \frac{e^{it \alpha}}{t} $: this, in general, does not converge, so that, to make clear what this means to be, you should probably state that you want to differentiate the operator $ f * \{ \text{p.v. } \frac{e^{it \alpha}}{t} \}$. Taking a closer look:

$$ \text{p.v. } \frac{e^{it\alpha}}{t} * f (x) = \lim_{\epsilon \to 0} \int_{|y-x|\ge \epsilon} \frac{f(y)e^{i(x-y)\alpha}}{x-y} dy \\ = \pi e^{i\alpha x} \mathcal{H}(e^{-iy\alpha}f) $$

Where $\mathcal{H}$ Denotes the Hilbert Transform (click here)

STEP 2: So, to differentiate this function, we should, somehow, differentiate the Hilbert Transform. One way to do this is to differentiate the tempered distribution $\text{p.v. } \frac{1}{t} $. We recall some basic concepts:

Let $f \in \mathcal{S} (\mathbb{R}) $, and $u \in \mathcal{S}'(\mathbb{R})$. We define $u' \in \mathcal{S}'(\mathbb{R})$ as the unique tempered distribution that satisfies

$$ u'(f) = u(f') \; \forall f \in \mathcal{S}(\mathbb{R})$$

From the alternative definition of the Hilbert Transform (i.e., $ (\mathcal{H}f)^{\wedge} = - i \;\text{sgn} \hat{f} $ ) we see that $ \frac{\text{d}}{\text{d}x} \mathcal{H}f ( x) = \mathcal{H} \left( \frac{\text{d}}{\text{d} x} f \right) (x) $, at least if $f \in \mathcal{S} ( \mathbb{R})$.

STEP 3: Looking back to our problem, we can, using the product rule to differentiate, get that

$$ \frac{\text{d}}{\text{d}x} \left( \text{p.v. } \frac{e^{i\alpha t}}{t} * f (x) \right) = \pi (i\alpha e^{i\alpha x} \mathcal{H}(e^{-i\alpha y} f) + e^{i\alpha x} \mathcal{H} ( e^{-iy\alpha}(-i\alpha f + f')) ) = \pi e^{ix \alpha} \mathcal{H} (e^{-iy\alpha}f') $$

Using some basic rules of the Fourier Transform, when we pass it on the last expression for $ f \in \mathcal{S} (\mathbb{R})$, we get something like

$$- \pi i \text{ sgn}(\theta - \frac{\alpha}{2\pi}) \hat{f'}(\theta) = - \pi i \cdot (2\pi i \theta) \text{ sgn}(\theta - \frac{\alpha}{2\pi}) \hat{f}(\theta) = 2 \pi^2 \theta \text{ sgn} (\theta - \frac{\alpha}{2\pi}) \hat{f}(\theta) $$

And this gives your desired Fourier Multiplier.

Remark 1: Note that this is almost the same as Cameron's result - maybe the reason is that, here, I am defining

$$ \hat{f} (\xi) = \int f(t) e^{-2\pi i t \xi} \text{d} t $$

And the fact that, in his arguing, he was looking at your operator, instead of the definition that I gave here.

Remark 2: Your original operator is likely not the distributional derivative of the Operator we are dealing with - in fact, if you look carefully for some smooth function $ f $ such that $\text{supp} f \subseteq [-1,1]$, $f\equiv 1 $ in a neighborhood of 1, then the second summand converges on the principal value sense, but the first one does not - and that makes impossible the study of the operator you wrote there. So, if one really wants to differentiate the way you did, the best way to do it is with the help of distributions and the fine properties of the Hilbert Transform.