Given any $g \in L^\infty(\mathbb{R})$, we define the associated multiplier operator $T_g \colon L^2(\mathbb{R}) \to L^2(\mathbb{R})$ by $$ \mathcal{F}(T_g f) \ = \ g.\mathcal{F}f $$ where $\mathcal{F}$ denotes the Fourier transform.

I am interested in knowing how far beyond $L^2(\mathbb{R})$ it is possible to take a function $f \colon \mathbb{R} \to \mathbb{C}$, and it still make at least some kind of sense to talk about a corresponding function $T_gf \colon \mathbb{R} \to \mathbb{C}$. Accordingly, I have tried to formulate the weakest definition of the "domain $\mathfrak{D}(g)$ of definition of $T_g$" that I can, within the constraint that I still wish to be in the world of functions, not distributions:

A sequence $\mathbf{r}\!=\!(r^{[k]})_{k \geq 1}$ of positive real numbers will be called a valid growth rate if there exists a test function $\phi \in C_c^\infty(\mathbb{R})$ such that $\phi(0)=\|\phi\|_\infty=1$ and for all $k \geq 1$, $\,\phi^{(k)}(0)=0$ and $\|\phi^{(k)}\|_\infty \leq r^{[k]}$.

Definition. For any $g \in L^\infty(\mathbb{R})$, let $\mathfrak{D}(g)$ be the set of all functions $f \in L_{\mathrm{loc}}^2(\mathbb{R})$ for which there exists a measurable function $\tilde{T}_{\!g}f \colon \mathbb{R} \to \mathbb{C}$, an increasing sequence of compact intervals $K_n \subset \mathbb{R}$ covering $\mathbb{R}$ and a sequence of valid growth rates $\mathbf{r}_n=(r_n^{[k]})$, such that the following holds: for every sequence of test functions $\phi_n \in C_c^\infty(\mathbb{R})$ with the properties that

  • $\phi_n=1$ on $K_n$ for each $n$,
  • $\|\phi_n\|_\infty=1$ and $\|\phi_n^{(k)}\|_\infty \leq r_n^{[k]}$ for each $n$ and $k$,

the sequence of functions $T_g(f\phi_n)$ converges in probability to $\tilde{T}_{\!g}f$ as $n \to \infty$.

The "convergence in probability" here is defined with respect to any probability measure on $\mathbb{R}$ that is equivalent to the Lebesgue measure. Alternatively, "convergence in probability" means that every subsequence admits a further subsequence converging Lebesgue-a.e. to the desired limit.

Remark. Formulating the definition in terms of convergence in probability (rather than something stronger like $L_{\mathrm{loc}}^p$-convergence) may be a bit risky, e.g. a sequence of zero-centred Gaussian PDFs whose standard deviation tends to zero converges in probability to the constant zero function.

Are there any existing results [or can we come up with some results] on how broad this class $\mathfrak{D}(g)$ is, for any particular class of functions $g$?
$\hspace{6mm}$For example: Is it the case that if $g$ is in some sense "sufficiently nice" (e.g. $g$ is smooth and each derivative has at most polynomial growth), then $\mathfrak{D}(g)$ includes the set of all bounded smooth functions all of whose derivatives are bounded?

Physical motivation: In signal processing, filters are typically multiplier operators, for which the multiplier $g$ is a rational function $P/Q$ where $\mathrm{order}(P) \leq \mathrm{order}(Q)$ and $Q$ has no real roots. Thus, in a sense, this question relates to the physical meaningfulness of filtering a signal which was recorded as just an extract from some more long-term process.

First thought: If $g$ is smooth and each derivative has at most polynomial growth, then $T_gf$ is a well-defined linear functional on $\mathcal{S}(\mathbb{R})$ for every $f \in L^\infty(\mathbb{R})$, by $$ T_gf(\bar{\varphi}) \ := \ \int_{\mathbb{R}} f(t)\overline{T_{\overline{g}}\varphi}(t) \, dt. $$ (I'm not sure if this fulfils the continuity requirement for being a tempered distribution.) In this case, I guess the question of whether $f \in \mathfrak{D}(g)$ can probably be reduced to whether Lebesgue-almost the whole real line can be covered by open sets $U$ for which $T_gf$ coincides with an $L_{\mathrm{loc}}^1$ function on $C_c^\infty(U)$.

  • 1
    $\begingroup$ What is the role of $\tau$? I guess that you want $\phi((t - \tau)/s)$ converging to $1$ but if $\phi(0) = 0$ when $s$ tends to infinity the expression goes to $0$ not to $T_g(f)$. $\endgroup$ Feb 18, 2018 at 12:21
  • $\begingroup$ I've now changed the definition, using what seems a more logical approach. $\endgroup$ Feb 19, 2018 at 20:44
  • $\begingroup$ I think that if $g$ is integrable and $\mathcal{F}g$ is integrable (e.g. if $g \in C^2$ with $g,g',g'' \in L^1$), then the convolution theorem gives an explicit formula for $T_gf \in L^1$ for all $f \in L^\infty$, and from there it will be essentially immediate that $L^\infty \subset \mathfrak{D}(g)$ and $\tilde{T}_gf=T_gf$ independently of the choice of test functions. But this does not solve the case described in the bounty. $\endgroup$ Feb 24, 2018 at 1:28

2 Answers 2


A worthy and celebrated extension is the Hörmander-Mihlin Theorem, which can be found in the first volume of The Analysis of Linear Partial Differential Operators, Grundlehren 256, Springer-Verlag: let me call $g(D)$ your operator $T_g$.

If $g$ is a bounded function on $\mathbb R^n$ such that $ |x|^{k}|\nabla ^{k}g| $is bounded for all $k\in [0, 1+n/2]$, then the operator $g(D)$ is bounded on $L^p$ for $1<p<\infty$. Moreover, thanks to the Marcinkiewicz Theorem it is also bounded from $L^1$ to weak $L^1$ (also from the Hardy space $\mathscr H^1$ into $L^1$ and from $L^\infty$ to BMO).

Moreover if $g$ is a "multiplier" of the Schwartz space, i.e. a smooth function increasing polynomially as well as all its derivatives, the operator $g(D)$ can be defined as an endomorphism of $\mathscr S'(\mathbb R^n)$: you define simply for $T\in \mathscr S'(\mathbb R^n)$, $$ \left(\text{Fourier}\left(g(D) T\right)\right)(\xi)=g(\xi)\hat T(\xi), $$ so that $$ \langle g(D) T, \phi\rangle_{\mathscr S'(\mathbb R^n),\mathscr S(\mathbb R^n)} = \langle \hat T, g\check{\hat\phi}\rangle_{\mathscr S'(\mathbb R^n),\mathscr S(\mathbb R^n)}, $$ where $ \check \psi(\xi)=\psi(-\xi). $ On the other hand, if $g$ is only bounded, $g(D)$ can be defined on $\mathscr S(\mathbb R^n)$ and extended as an endomorphism of $L^2(\mathbb R^n)$; above I give an example of what happens if you control more derivatives : you obtain some $L^p$ boundedness. So the "maximal" extension to $\mathscr S'(\mathbb R^n)$ requires full smoothness and temperate growth whereas $L^2$ extension can be done with $g$ only bounded and $L^p$ extension requires a specific behaviour for a finite number of derivatives.

  • $\begingroup$ Thanks for this. Presumably, when you say "thanks to the Marcinkiewicz Theorem", you are referring to some extrapolation theorem, not the standard Marcinkiewicz Interpolation Theorem? $\endgroup$ Feb 26, 2018 at 23:47
  • $\begingroup$ In fact I refer to the fact that the maximal function sends $L^1$ in $L^1_w$ and then I use the $L^2$ bounedness and the Marcinkiewicz Interpolation Theorem to get the $L^p$ boundedness for $p\in (1,2]$ and duality for finite exponents. $\endgroup$
    – Bazin
    Feb 27, 2018 at 14:32
  • $\begingroup$ Okay thanks; but how do you get to boundedness from $L^\infty$ to BMO? (This is really the one case I am most interested in.) $\endgroup$ Feb 27, 2018 at 15:49
  • $\begingroup$ You get the boundedness from the Hardy space $\mathscr H^1$ into $L^1$ with the atomic decomposition and by duality follows the boundedness from $L^\infty$ to BMO (which is the dual of $\mathscr H^1$). $\endgroup$
    – Bazin
    Feb 27, 2018 at 19:04

I think I can now prove the following (which covers the case requested in the bounty):

Theorem. Let $g=P/Q$ for polynomials $P$ and $Q$ where $\mathrm{order}(P) \leq \mathrm{order}(Q)$ and $Q$ has no real roots. Then $\mathfrak{D}(g)$ includes all $L_{\mathrm{loc}}^2(\mathbb{R})$ functions $f$ for which $\frac{1}{t}\log|f(t)| \to 0$ as $t \to \pm\infty$.

Proof. Express $g$ in partial fractions as $$ g(x) \ = \ \alpha_0 + \sum_{k=1}^N \frac{\alpha_k}{(x-(a_k + ib_k))^{n_k}} $$ where $\alpha_k \in \mathbb{C}$, $a_k \in \mathbb{R}$, $b_k \in \mathbb{R}\!\setminus\!\{0\}$, and $n_k \in \mathbb{Z}_{\geq 1}$. For each $1 \leq k \leq N$, let $$ g_k(x) \ = \ \frac{1}{(x-(a_k + ib_k))^{n_k}}. $$ Then $g_k \in L^2$ and it is known that $$ \mathcal{F}^{-1}g_k(t) \ = \ \left\{ \begin{array}{c l} \ \,\, C_k \mathbf{1}_{[0,\infty)\!}(t) \, t^{n_k-1}e^{2\pi(-b_k+ia_k)t} & b_k>0 \\ C_k \mathbf{1}_{(-\infty,0]\!}(t) \, t^{n_k-1}e^{2\pi(-b_k+ia_k)t} & b_k<0 \end{array} \right. $$ for some constant $C_k$. In particular, $h_k$ is bounded and decays exponentially fast. Hence, the function $h:=\mathcal{F}^{-1}(g-\alpha_0)$ is bounded and decays exponentially fast.

For any compactly supported $L^2$ function $\phi \colon \mathbb{R} \to \mathbb{C}$, the convolution theorem gives that $$ T_g\phi(t) \ = \ \alpha_0\phi(t) + \int_\mathbb{R} \phi(s)\,h(t-s) \, ds. $$ [Note in particular that if all the poles of $g$ have positive imaginary part, then $T_g\phi(t)$ depends only on $(\phi(s):s \leq t)$; presumably this is why in such a case, the filter $T_g$ is called a causal filter.]

Now fix any $t \in \mathbb{R}$. For any $f \in L_{\mathrm{loc}}^2(\mathbb{R})$ with $\frac{1}{s}\log|f(s)| \to 0$ as $s \to \pm\infty$, we have that $f(s)h(t-s)$ decays exponentially as $s \to \pm\infty$, and so is integrable in $s$. Therefore, by the dominated convergence theorem, for any sequence of test functions $\phi_n$ with $\|\phi_n\|_\infty=1$ converging pointwise to $1$, we have that $$ \hspace{40mm} T_g(f\phi_n)(t) \,\to\, \alpha_0f(t) + \int_\mathbb{R} f(s)h(t-s) \, ds. \hspace{38mm} \square $$

In the last step, it should be quite easy to show that if $\phi_n$ converges uniformly on compact sets to $1$ (as in the definition of $\mathfrak{D}(g)$), then $T_g(f\phi_n)$ converges uniformly on compact sets to $\tilde{T}_{\!g}f$.

  • $\begingroup$ I will still give the bounty to anyone who gives interesting further results on the general question. $\endgroup$ Feb 24, 2018 at 19:00

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