V. Ya. Lin

ON EQUIVALENT NORMS IN THE SPACE OF FOURIER TRANSFORMS OF FINITE FUNCTIONS

(Presented by Academician L. S. Pontryagin on 23 XII 1961)

1. Let \(R^\nu\) and \(R_\nu\) be \(\nu\)-dimensional real Euclidean spaces of vectors
   \(x=(x^1,x^2,\ldots,x^\nu)\) and \(\xi=(\xi_1,\xi_2,\ldots,\xi_\nu)\); let \(\Omega\) be a measurable bounded set in \(R^\nu\), \(\operatorname{mes}(\Omega)>0\). Denote by \(\hat L_2(\Omega)\) the space of Fourier transforms \(\hat u(\xi)\) of complex functions \(u(x)\) equal to zero outside \(\Omega\) and square summable, with the natural norm

\[ \|\hat u\|=\left(\int_{R_\nu}|\hat u(\xi)|^2\,d\xi\right)^{1/2} =\left(\int_{\Omega}|u(x)|^2\,dx\right)^{1/2}=\|u\|. \tag{1} \]

Let \(M\) be a measurable set in \(R_\nu\), \(0<\operatorname{mes}(M)\leq \infty\); \(\mathfrak M=R_\nu\setminus M\),
\(E_\Omega=\{\hat u:\hat u\in \hat L_2(\Omega),\ \|\hat u\|=1\}\). Put

\[ \|\hat u\|_{M}=\left(\int_M|\hat u(\xi)|^2\,d\xi\right)^{1/2},\qquad \gamma_\Omega(\mathfrak M)=\sup_{\hat u\in E_\Omega}\int_{\mathfrak M}|\hat u(\xi)|^2\,d\xi . \]

The norms \(\|\cdot\|_M\) and \(\|\cdot\|\) are equivalent in \(\hat L_2(\Omega)\) if and only if
\(\gamma_\Omega(\mathfrak M)<1\). Following B. P. Paneiah \((^1)\), we shall call the set \(\mathfrak M\) determining if \(\gamma_\Omega(\mathfrak M)<1\) for every bounded \(\Omega\subset R^\nu\). In the work of B. P. Paneiah \((^2)\), for \(\nu=1\) the problem of describing determining sets is completely solved, while for \(\nu>1\) sufficient conditions are given. In the present note another approach to the problem is proposed, which for \(\nu>1\) makes it possible to obtain sufficient conditions different from those contained in \((^{1,2})\).

1. Consider in \(\hat L_2(\Omega)\) the bilinear functional

\[ \Phi_{\mathfrak M}(\hat u,\hat v)=\int_{\mathfrak M}\hat u(\xi)\,\overline{\hat v(\xi)}\,d\xi, \]

which is evidently positive, symmetric, and bounded. It is clear that

\[ \sup_{\hat u\in E_\Omega}\Phi_{\mathfrak M}(\hat u,\hat u)=\gamma_\Omega(\mathfrak M). \]

Consequently, there exists a positive self-adjoint bounded operator \(A_{\mathfrak M}\), acting in \(\hat L_2(\Omega)\), such that

\[ \Phi_{\mathfrak M}(\hat u,\hat v)=(A_{\mathfrak M}\hat u,\hat v), \qquad \|A_{\mathfrak M}\|_{\hat L_2(\Omega)}=\gamma_\Omega(\mathfrak M). \]

It is easy to show that if \(F\) and \(F^{-1}\) are the direct and inverse Fourier operators, and \(\chi_{\mathfrak M}\) and \(\chi_\Omega\) are the operators of multiplication by the characteristic functions of the sets \(\mathfrak M\) and \(\Omega\)*, then

\[ A_{\mathfrak M}\hat u =F\chi_\Omega F^{-1}\chi_{\mathfrak M}\hat u =\frac{1}{(2\pi)^\nu}\int_{\mathfrak M}K_\Omega(\xi-\eta)\hat u(\eta)\,d\eta, \tag{2} \]

where

\[ K_\Omega(\xi)=\int_\Omega e^{i(x,\xi)}\,d\xi . \tag{3} \]

* The operators \(\chi_{\mathfrak M}\) and \(\chi_\Omega\) act, respectively, from \(\hat L_2(\Omega)\) to \(L_2(\mathfrak M)\) and from \(\hat L_2(\mathfrak M)\) to \(L_2(\Omega)\), where \(\check L_2(\mathfrak M)\) is the space of inverse Fourier transforms of functions from \(L_2(\mathfrak M)\).

Definition. We shall call a set \(\mathfrak M \subset R_\nu\) a set of attainability if, for every bounded \(\Omega \subset R^\nu\), there exists a function \(\hat u_\Omega \in E_\Omega\) such that
\[ \gamma_\Omega(\mathfrak M)=\Phi_{\mathfrak M}(\hat u_\Omega,\hat u_\Omega). \]

Lemma 1. Every set of attainability whose complement has positive measure is determining.

Lemma 2. In order that \(\mathfrak M\) be a set of attainability, it is necessary and sufficient that, for every bounded \(\Omega \subset R^\nu\), the operator \(A_{\mathfrak M}\) have a maximal vector in \(\hat L_2(\Omega)\).

Theorem 1. If, for every bounded \(\Omega \subset R^\nu\),
\[ \int_{\mathfrak M \times \mathfrak M} |K_\Omega(\xi-\eta)|^2\,d\xi\,d\eta < \infty, \tag{4} \]
then \(\mathfrak M\) is a set of attainability and is determining.

Proof. Consider, along with the operator \(A_{\mathfrak M}\), the operator \(\widetilde A_{\mathfrak M}=\chi_{\mathfrak M}A_{\mathfrak M}\), acting from \(\hat L_2(\Omega)\) into \(L_2(\mathfrak M)\). Using formula (2), the operators \(A_{\mathfrak M}\) and \(\widetilde A_{\mathfrak M}\) can be extended from \(\hat L_2(\Omega)\) to \(L_2(R_\nu)\). Here \(A_{\mathfrak M}\) maps \(L_2(R_\nu)\) into \(\hat L_2(\Omega)\), while \(\widetilde A_{\mathfrak M}\) maps \(L_2(R_\nu)\) into \(L_2(\mathfrak M)\). It follows from (2) that \(\widetilde A_{\mathfrak M}\) is an integral operator with kernel \(\chi_{\mathfrak M}(\xi)K_\Omega(\xi-\eta)\), and since the range of \(\widetilde A_{\mathfrak M}\) is contained in \(L_2(\mathfrak M)\), (4) implies the complete continuity of \(\widetilde A_{\mathfrak M}\). It is easy to see that
\[ A_{\mathfrak M}^2=A_{\mathfrak M}\widetilde A_{\mathfrak M}, \]
and hence the operator \(A_{\mathfrak M}^2\) is completely continuous. The operator \(A_{\mathfrak M}\) is self-adjoint on \(\hat L_2(\Omega)\) and, consequently, is completely continuous together with \(A_{\mathfrak M}^2\). Therefore it has a maximal vector. It remains to note that (4) implies \(\operatorname{mes}(R_\nu\setminus\mathfrak M)=\infty\), and to use Lemmas 2 and 1.

Remark. Theorem 1 remains valid if condition (4) is replaced by the condition
\[ \int_{\mathfrak M \times \mathfrak M}\prod_{i=1}^{\nu} \frac{\sin^2 \tau(\xi_i-\eta_i)}{(\xi_i-\eta_i)^2}\,d\xi\,d\eta < \infty \tag{5} \]
for some \(\tau>0\).

Let \(\check L_2(\mathfrak M)\) be the space of inverse Fourier transforms \(\check u(x)\) of functions \(u(\xi)\in L_2(\mathfrak M)\), with the natural norm
\[ \|\check u\|=\left(\int_{R^\nu}|\check u(x)|^2\,dx\right)^{1/2}, \]
and let
\[ E_{\mathfrak M}=\{\check u:\check u\in\check L_2(\mathfrak M),\ \|\check u\|=1\}. \]
Put
\[ \gamma_{\mathfrak M}(\Omega)=\sup_{\check u\in E_{\mathfrak M}}\int_\Omega |\check u(x)|^2\,dx. \]

Theorem 2. The equality
\[ \gamma_{\mathfrak M}(\Omega)=\gamma_\Omega(\mathfrak M) \]
holds.

Proof. Introduce the operator
\[ A_\Omega=F^{-1}\chi_{\mathfrak M}F\chi_\Omega, \]
acting in \(\check L_2(\mathfrak M)\). Then the following relations are valid:
\[ \gamma_{\mathfrak M}(\Omega)=\|F^{-1}\chi_{\mathfrak M}F\chi_\Omega\|_{\check L_2(\mathfrak M)} \le \|\chi_{\mathfrak M}\|_{\hat L_2(\Omega)}\cdot \|\chi_\Omega\|_{\check L_2(\mathfrak M)} = \sqrt{\gamma_\Omega(\mathfrak M)}\cdot \sqrt{\gamma_{\mathfrak M}(\Omega)}, \]
\[ \gamma_\Omega(\mathfrak M)=\|F\chi_\Omega F^{-1}\chi_{\mathfrak M}\|_{\hat L_2(\Omega)} \le \|\chi_\Omega\|_{\check L_2(\mathfrak M)}\cdot \|\chi_{\mathfrak M}\|_{\hat L_2(\Omega)} = \sqrt{\gamma_{\mathfrak M}(\Omega)}\cdot \sqrt{\gamma_\Omega(\mathfrak M)}. \]
The rest is obvious.

Theorem 2 makes it possible, for every assertion about determining sets, to formulate a dual assertion. As an example, we state the following theorem, which is dual to the theorem of the paper \((^2)\):

Theorem 3. Let \(\Omega\) be a set in \(R^1\), not necessarily bounded, with \(\beta(\Omega)<1\)*, and let \(\mathfrak M\) be any bounded set in \(R_1\). Then \(\gamma_\Omega(\mathfrak M)<1\), and consequently the norms \(\|\cdot\|_M\) and \(\|\cdot\|\) are equivalent in \(\hat L_2(\Omega)\).

\[ \text{* For the definition of the quantity } \beta \text{ see } (^{1,2}). \]

1. Here we shall give one more sufficient condition for the equivalence of the norms \(\|\cdot\|_M\) and \(\|\cdot\|\) in \(\hat L_2(\Omega)\), not connected with consideration of the operator \(A_{\mathfrak M}\).

Let \(h>0,\ a\in R_\nu,\ K_h^a=\{\xi:\ |\xi_i-a_i|\le h/2,\ i=1,\ldots,\nu\},\ K_N=\{\xi:\ |\xi_i|\le N\}\). Put

\[ \alpha(\mathfrak M)=\lim_{h\to 0}\left\{\lim_{N\to\infty}\left[\sup_{a\in K_N}\frac{\operatorname{mes}(\mathfrak M\cap K_h^a)}{h^\nu}\right]\right\}. \tag{6} \]

It is clear that for any measurable \(\mathfrak M\subset R_\nu\), \(0\le \alpha(\mathfrak M)\le 1\).

Theorem 4. If \(\alpha(\mathfrak M)<1\), then the norms \(\|\cdot\|_M\) and \(\|\cdot\|\) are equivalent in \(\hat L_2(\Omega)\) for every bounded \(\Omega\subset R^\nu\), and, consequently, \(\mathfrak M\) is a determining set.

We precede the proof of Theorem 4 by two lemmas.

Lemma 3. If for all \(\hat u\in \hat L_2(\Omega)\)

\[ \|\hat u\|^2\le C_1\int_K|\hat u(\xi)|^2\,d\xi+C_2\int_M|\hat u(\xi)|^2\,d\xi, \]

where \(K,M\subset R_\nu\), with \(K\) bounded; \(\operatorname{mes}(M)>0\); \(C_1,C_2\) are constants not depending on \(\hat u\), then the norms \(\|\cdot\|_M,\ \|\cdot\|\) are equivalent in \(\hat L_2(\Omega)\).

Let \(m=(m_1,m_2,\ldots,m_\nu)\) be a \(\nu\)-dimensional index running through all nodes of the \(\nu\)-dimensional integer lattice, \(\Delta_m^h=\{\xi:\ |\xi_i-m_i h|\le h/2,\ i=1,2,\ldots,\nu\}\), \(\Omega_\tau=\{x:\ |x^i|\le \tau,\ i=1,2,\ldots,\nu\}\).

Lemma 4. There exists a positive function \(C_\nu(\sigma)\), defined for \(\sigma>0\) and tending to zero together with \(\sigma\), such that, when \(C_\nu(\tau h)<1\), for all \(\hat u\in \hat L_2(\Omega_\tau)\) the inequality

\[ \|\hat u\|^2\le \frac{h^\nu}{1-C_\nu(\tau h)}\sum_m\inf_{\xi\in\Delta_m^h}|\hat u(\xi)|^2 \]

holds.

For example, one can show that the function \(C_\nu(\sigma)=\nu^{3/2}\sigma(1+\sigma/\pi)^\nu e^{\nu\sigma}\) has all the indicated properties.

We now prove Theorem 4. Obviously, it is enough to prove that for arbitrary \(\tau>0\) the norms \(\|\cdot\|_M\) and \(\|\cdot\|\) are equivalent in \(\hat L_2(\Omega_\tau)\). Fix an arbitrary \(\tau_0>0\) and choose \(h_0>0\) so small that, for \(h<h_0\), the inequality \(C_\nu(\tau_0 h)<1\) holds. From the condition \(\alpha(\mathfrak M)<1\) it follows that there exist numbers \(\delta>0,\ h<h_0\), and \(N>0\) such that, for \(a\in K_N\), the inequality \(\operatorname{mes}(\mathfrak M\cap K_h^a)\le h^\nu(1-\delta)\) holds. Taking into account that \(M=R_\nu\setminus \mathfrak M\), we conclude that \(\operatorname{mes}(M\cap K_h^a)\le \delta h^\nu\) for \(a\in K_N\). Let \(a_m\in R_\nu\) be the point with coordinates \((m_1h,m_2h,\ldots,m_\nu h)\), \(K_m=K_h^{a_m}\). Taking into account that \(\Delta_m^h=K_h^{a_m}=K_m\), and applying Lemma 4, we obtain

\[ \|\hat u\|^2\le \frac{h^\nu}{1-C_\nu(\tau_0h)}\sum_m\inf_{\xi\in K_m}|\hat u(\xi)|^2= \]

\[ =\frac{h^\nu}{1-C_\nu(\tau_0h)} \sum_{m:\,a_m\in K_N}\inf_{K_m}|\hat u(\xi)|^2 +\frac{h^\nu}{1-C_\nu(\tau_0h)} \sum_{m:\,a_m\notin K_N}\inf_{K_m}|\hat u(\xi)|^2\le \]

\[ \le \frac{1}{1-C_\nu(\tau_0)} \sum_{m:\,a_m\in K_N}\int_{K_m}|\hat u(\xi)|^2\,d\xi+ \]

\[ +\frac{1}{1-C_\nu(\tau_0 h)} \sum_{m:\, a_m \in K_N} \frac{h^\nu}{\operatorname{mes}(M\cap K_m)} \,\operatorname{mes}(M\cap K_m)\inf_{K_m}|\hat u(\xi)|^2 \leq \]

\[ \leq \frac{1}{1-C_\nu(\tau_0 h)} \int_{K_{N+h/2}} |\hat u(\xi)|^2\,d\xi + \frac{1}{\delta[1-C_\nu(\tau_0 h)]} \sum_m \operatorname{mes}(M\cap K_m)\inf_{K_m}|\hat u(\xi)|^2 \leq \]

\[ \leq \frac{1}{1-C_\nu(\tau_0 h)} \int_{K_{N+h/2}} |\hat u(\xi)|^2\,d\xi + \frac{1}{\delta[1-C_\nu(\tau_0 h)]} \int_M |\hat u(\xi)|^2\,d\xi, \]

and to complete the proof it remains only to use Lemma 3.

In a similar way the following theorem can be proved:

Theorem 5. Let \(\mathfrak M_1\) be a determining set, and let \(\alpha(\mathfrak M_2)=0\). Then the set \(\mathfrak M=\mathfrak M_1\cup\mathfrak M_2\) is determining.

Examples of sets \(\mathfrak M\) for which \(\alpha(\mathfrak M)=0\) may be all sets of finite measure, but not only these. We note that in Theorem 5 the condition \(\alpha(\mathfrak M_2)=0\) cannot be replaced by the condition \(\alpha(\mathfrak M_2)<1\).

Received
23 XI 1961

REFERENCES

\(^{1}\) B. P. Paneiakh, DAN, 138, No. 1 (1961).
\(^{2}\) B. P. Paneiakh, DAN, 142, No. 5 (1962).