Context. I am trying to undestand the theory underlying "Beppo-Levi"-like spaces defined as

$$ H = \left\{f\in {\cal S}'(\mathbb{R}^d) \;\left| \; t\times\widetilde{f} \in {\cal L}^2(\mathbb{R}^d) \right.\right\} $$

where $\widetilde{f}$ is the Fourier transform of the tempered distribution $f$, and $t(\xi)$ is a smooth radial function (i.e., a $C^\infty$ function of $|\xi|$), such that $1/|t|^2$ is integrable at infinity, and everywhere non-zero except in $\xi=0$ where one has $t(\xi)\sim C|\xi|^m$ for some order $m\geq 1$.

This is a direct generalization of the Beppo-Levi spaces which underlie thin-plate splines interpolation, as described e.g. by Duchon [1], Meinguet [2], Wahba [3], etc., and that correspond to taking $t(\xi)=|\xi|^m$ (with $m>d/2$ to ensure square-integrability at infinity). Just like the original Beppo-Levi spaces, $H$ contains all distributions $f\in{\cal S}'$ such that $t\times \widetilde{f}=0$, which turns out to be the space ${\cal P}_{m-1}$ of polynomials with degree $<m$.

Question. I am trying to prove that the elements of H are continuous functions. Using the Fourier inversion formula, one can show easily the following result : For every $f\in H$, if some distribution $T=\sum_{j=1}^n c_j \delta_{x_j}$ is such that $\langle T,p\rangle = 0$ for every $p\in {\cal P}_{m-1}$, then $$ T*f = \sum_{j=1}^n c_j \tau_{x_j}f $$ is a continuous function. Is this result sufficient to prove that $f$ itself is a function, moreover continous?

Indeed, for almost every point set $\{x_j\}_{j=1\dots n}$ with sufficiently many points, one can find appropriate weights $c_j$ such that distribution $T=\sum_j c_j\delta_{x_j}$ annihilates all polynomials in ${\cal P}_{m-1}$. So intuitively, I would say that the continuity of all such functions $T*f$ can only be achieved if $f$ itself is a function, moreover continuous. However, I fail to find a rigorous argument.

Remarks. This is the approach followed by Meinguet [2] in his study of the original Beppo-Levi space ($t(\xi)=|\xi|^m$). He proves the continuity of $f$ thanks to the above result, and a smart choice of distribution $T$. However, unless I misread his argument, he seems to assume without proof that the elements of $H$ are indeed functions, so that expressions like ``$f(x_j)$'' make sense.

EDIT: To make my problem clearer, let me summarize Meinguet's argument below.

Letting $J:={\rm dim}({\cal P}_{m-1})$, he considers a set of points $\{a_j\}_{j=1\dots J}$ that is unisolvant over ${\cal P}_{m-1}$, meaning that for every possible values $(\beta_j)\in\mathbb{R}^J$, there is a unique polynomial $p\in{\cal P}_{m-1}$ solution to the linear system $\forall j,\; p(a_j)=\beta_j$. Let $\{p_j\}$ in ${\cal P}_{m-1}$ the dual polynomials to the points $\{a_j\}$, i.e., such that $p_j(a_i)=\delta_{ij}$. He then introduces, for every $x\in\mathbb{R}^d$, the distribution

$$ T_{(x)} := \delta_x - \sum_{j=1}^J p_j(x)\delta_{a_j} $$

One verifies instantly that $\langle T_{(x)}, p \rangle=0$ for all $p\in{\cal P}_{m-1}$, so that the above property is verified : for every $f\in H$, $T_{(x)}* f$ is a continuous function.

Moreover, using the semi-Hilbert structure of space $H$ and the existence of a reproducing kernel, he argues that for any sequence $(x_n)$ such that $x_n \to x$, one has

$$ T_{(x_n)}*f \to T_{(x)}*f $$

as continuous functions, the convergence being uniform over any compact set.

From that, he concludes directly that $f(x_n)\to f(x)$, i.e., $f$ is a continuous function. This is the part of the argument that I do not understand. Isn't this argument valid only if one presupposes that $f$ is indeed a function, so that expressions $f(x_n)$ and $f(x)$ actually make sense??

[1] Duchon, Jean, Splines minimizing rotation-invariant semi-norms in Sobolev spaces, Constr. Theory Funct. several Variables, Proc. Conf. Oberwolfach 1976, Lect. Notes Math. 571, 85-100 (1977). ZBL0342.41012.

[2] Meinguet, Jean, An intrinsic approach to multivariate spline interpolation at arbitrary points, Polynomial and spline approximation, Proc. NATO adv. Study Inst., Calgary 1978, NATO adv. Study Inst. Ser., Ser. C - math. phys. Sci. 49, 163-190 (1979). ZBL0413.41007.

[3] Wahba, Grace, Spline models for observational data, CBMS-NSF Regional Conference Series in Applied Mathematics. 59. Philadelphia, PA: SIAM, Society for Industrial and Applied Mathematics. XII, 169 p. (1990). ZBL0813.62001.

  • $\begingroup$ Note that if $f$ is a function and $\sum c_j \tau_{t_j} f$ is continuous, you cannot conclude that $f$ is continuous. The simplest counterexample is a discontinuous periodic function $f$, with $t_1,t_2$ periods and $c_1+c_2=0$. The characteristic function of $\mathbb Q$ is actually a very ugly example.... $\endgroup$
    – Nick S
    Commented Aug 26, 2022 at 22:36
  • $\begingroup$ Indeed. However here we have a stronger result that $\sum c_j \tau_{x_j}f$ is continuous for any choice of offsets {$x_j$} and weights {$c_j$} such that $\sum c_j p(x_j)=0$ for all polynomials $p\in{\cal P}_{m-1}$. In particular, almost every set of offsets $\{x_j\}$ with more elements that ${\rm dim}{\cal P}_{m-1}$ can be used in the property, provided they are associated to the correct weights $c_j$. So I can hardly imagine how to build a counterexample function $f$ that would not be continuous and still verify the property. $\endgroup$ Commented Aug 27, 2022 at 8:45

1 Answer 1


After some more thought, I gave up Meinguet's approach above (based on sums of the form $\sum_j c_j\tau_{x_j}f$) for a more direct approach in order to prove continuity of the elements of $H$. Any feedback or criticism is welcome.

Note as above

$$ H = \left\{f\in {\cal S}'(\mathbb{R}^d) \;\left| \; t\times\widetilde{f} \in {\cal L}^2(\mathbb{R}^d) \right.\right\} $$

Now introduce any smooth window function $k(\xi)$ such that $0\leq|k|\leq 1$, with $k(\xi)=1$ for $|\xi|\leq 1/2$ and $k(\xi)=0$ for $|\xi|\geq 1$. Define the new weighting function

$$ t_k(\xi):= k(\xi) + (1-k(\xi))t(\xi) $$

that behaves like $t$ at infinity, but converges to a nonzero constant in $\xi=0$ with $t_k(0)=k(0)=1$. Assume also that $t_k$ is everywhere nonzero (this is always possible by modifying function $k$). Hence, function $1/t_k$ is well-defined and belongs to ${\cal L}^2$ (given our hypotheses on $t$). By a classic argument, this implies that the modified function space associated to the weighting function $t_k$:

$$ H_k := \left\{f\in {\cal S}'(\mathbb{R}^d) \;\left| \; t_k\times\widetilde{f} \in {\cal L}^2(\mathbb{R}^d) \right.\right\} $$

is a Reproducing Kernel Hilbert Space (RKHS), with elements of the form $f\in H_k \Longleftrightarrow f=K_k*u$ with $K_k\in{\cal L}^2$ the inverse Fourier transform of $1/t_k$, and $u$ any function in ${\cal L}^2$. In particular, such $f\in H_k$ is always a continuous function.

Going back to the original function space $H$, any $f\in H$ can be decomposed as $f=f_1+f_2$ such that

$$ \widetilde{f_1} = k.\widetilde{f} \quad{\rm and}\quad \widetilde{f_2}= (1-k).\widetilde{f} $$

Then, $f_2$ is easily shown to belong to $H_k$ above, and thus, is a continuous function. On the other hand, $\widetilde{f_1}$ is a distribution with compact support, so by a classic result its inverse Fourier transform $f_1$ is a $C^\infty$ function. Hence, the sum $f=f_1+f_2$ is indeed a continuous function.


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