I was recently reading about the Mikhlin and Hörmander Multiplier Theorems, which give conditions for a measurable function $m:\mathbb R^d\to\mathbb C$ to be an $L^p$ multiplier, i.e. for there to exist some $C_p$ such that for all $f\in\mathscr{S}(\mathbb R^d)$, $$\lVert T_m(f)\rVert_{L^p} = \lVert (m\hat{f})^\vee\rVert_{L^p}\le C_p\lVert f\rVert_{L^p}$$ which typically require some degree of $C^k$-regularity on $m$.

However, it is not clear to me in what sense the operator $T_m$ can be extended by density. Of course, this can be done formally by Hahn-Banach, but for $f\in L^p$ for $2<p<\infty$ (or even $p=\infty$), do we know that $\hat f\in\mathscr{S}'(\mathbb R^d)$ will be sufficiently regular that $m\hat f$ can actually be defined for, say, $m\in C^d(\mathbb R^d)$? Put more succinctly, what is the worst possible order of the tempered distribution $\hat f$ for $f\in L^p(\mathbb R^d)$, where $2<p\le\infty$? According to this question on MO, in general the distribution $\hat f$ may be of positive order, but I have yet to find further results in this direction.


You do not need Hahn-Banach to extend a continuous linear map defined in a dense subspace of a Hausdorff topological vector space into another, complete Hausdorff topological vector space. The extension is uniquely defined by (uniform) continuity and even satisfies the same (semi)norm bounds.

In raw detail, one simply defines the image of the limit of a Cauchy net in the dense subspace as the limit of the image of that net. Uniform continuity ensures that this new net is also Cauchy, and continuous linear maps are always uniformly continuous (with respect to the uniform structures induced by the vector topologies in the domain and codomain). Completeness and the Hausdorff property of the codomain, on their turn, guarantee that the latter limit exists and is unique. It is straightforward to verify that linearity and preservation of (semi)norm bounds by the extension hold.

Because of that, it does not matter if the multiplier representation of $T_m$ only makes sense in the dense subspace $\mathscr{S}$ (endowed with the $L^p$ norm). In fact, even the Fourier transform itself is defined in (say) $L^2$ like this - strictly speaking, the classical Fourier integral formula for $\hat{f}$ does not make sense for a general $f\in L^2$ but only in the dense subspaces $\mathscr{S}$ or $L^2\cap L^1$. Nevertheless, thanks to the Plancherel formula, the linear map defined by this formula extends uniquely and continuously to $L^2$ (but not the formula itself). The same is true of more general Fourier integral operators.

Since the Fourier transform is a topological isomorphism of $\mathscr{S}'$ onto itself and all $L^p$ spaces embed into $\mathscr{S}'$, one can even turn this reasoning around and define $m\hat{f}$ as the Fourier transform of $T_m f$ for $f\in L^p$, $1\leq p\leq\infty$, with $T_m$ uniquely defined by continuous extension from the dense subspace $\mathscr{S}$ through the $L^p$ bounds you wrote. In other words, if $T_m$ is bounded in the $L^p$ norm, it makes sense to define $m\hat{f}$ like this for any $f\in L^p$. The whole point of the multiplier theorems is to single out classes of $m$'s for which this operation makes sense.

Edit: the above argument addresses the OP's concern about whether it is needed to make actual sense of $m\hat{f}$ for $f\in L^p$ in order to define $T_m$. That being said, let us now assess the question in the title, which was not done before. Theorem 7.9.3, pp. 242 of L. Hörmander's book The Analysis of Linear Partial Differential Operators I - Distribution Theory and Fourier Analysis (Springer-Verlag, 1983) shows that if $f\in L^p(\mathbb{R}^n)$, then $\hat{f}$ belongs to the (negative-order) $L^2$ Sobolev space $H_{-s}(\mathbb{R}^n)$ for $2<p\leq\infty$ and $s>n(\frac{1}{2}-\frac{1}{p})$, and $H_{-s}$ consists of distributions of order $k$ for $s\leq k\in\mathbb{N}$. Conversely, Theorem 7.6.6, pp 209-210 of the same book shows that if the Fourier transform of $L^p(\mathbb{R}^n)$ consists of distributions of order $k$, then $k\geq n(\frac{1}{2}-\frac{1}{p})$. In other words, there certainly are distributions of order greater but not smaller (ahem) than $n(\frac{1}{2}-\frac{1}{p})$ which are Fourier transforms of elements of $L^p(\mathbb{R}^n)$ if $p>2$.

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    $\begingroup$ That’s a bit besides the point of my question though. I was wondering specifically about just how “rough” the Fourier transform of a bounded function can get. $\endgroup$ – D. Wynter Jan 10 '18 at 0:27
  • $\begingroup$ @MonstrousMoonshine: I agree, but you did (unfortunately) use the Mikhlin-Hormander theorem as a motivating example. It may pay to edit the question so that the motivation is somewhat de-emphasized. $\endgroup$ – Willie Wong Jan 10 '18 at 2:26
  • $\begingroup$ Actually, I’ve found a potential issue in your answer — $m$ need not be smooth, according to the Mikhlin Multiplier Theorem. Therefore an upper bound on the order is in fact required to do what you’re suggesting. $\endgroup$ – D. Wynter Jan 10 '18 at 2:34
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    $\begingroup$ I've added the missing discussion on the title's question. Moreover, any $m\in\mathscr{S}′$ yields $T_m$ as a map from $\mathscr{S}$ into $\mathscr{S}′$, since it's just convolution with $\check{m}$ which also belongs to $\mathscr{S}′$. Since $L^\infty\subset\mathscr{S}'$, any Fourier multiplier fits into this picture. In other words, it makes sense to ask whether $T_m$ maps $\mathscr{S}$ into $L^p$ for a given $m\in L^\infty$ and, if that's the case, whether it's bounded in the $L^p$ norm. $\endgroup$ – Pedro Lauridsen Ribeiro Jan 10 '18 at 3:12

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