Abstract Generated abstract
This paper studies completeness of the system of all translates of a function in weighted spaces whose norm involves an even subadditive weight, and relates this problem to regularity of the corresponding convolution normed ring. It proves that if the weight satisfies an integrability condition, then the ring is regular and the translates of a function are complete exactly when its Fourier transform has no real zeros. For sublinear weights satisfying monotonicity assumptions but failing this integrability condition, the paper shows that the Fourier transforms form a quasi-analytic class and the ring is nonregular. It also derives necessary completeness conditions analogous to Korenblum’s conditions for exponential weights, expressed through the inverse of an integral determined by the weight.
Full Text
Reports of the Academy of Sciences of the USSR
1967. Volume 173, No. 4
UDC 517.522 + 519.48
MATHEMATICS
S. P. GEISBERG
COMPLETENESS OF TRANSLATES OF A FUNCTION IN CERTAIN SPACES
(Presented by Academician V. I. Smirnov on 28 V 1966)
1°. The completeness* of the system of functions \(\{f_x(\tau)\}\), \(f_x(\tau)=f(x+\tau)\), \(-\infty<x<\infty\), in the spaces \(L_\alpha\),
\[ \|f\|=\int_{-\infty}^{\infty}|f(\tau)|e^{\alpha(\tau)}\,d\tau, \]
where \(\alpha(\tau)=\alpha(-\tau)>0\), \(\alpha(\tau)\) is nondecreasing, \(\alpha(\tau)/\tau\) is nonincreasing, \(\alpha(\tau_1+\tau_2)\leq \alpha(\tau_1)+\alpha(\tau_2)\), was studied in \((^{1-3})\). It turned out that the conditions for completeness depend on the rate of growth of \(\alpha(\tau)\). Shilov \((^2)\) showed that, for \(\alpha(\tau)=o(\ln \tau)\), the system \(\{f_x(\tau)\}\) is complete in \(L_\alpha\) if and only if
\[ \tilde f(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} f(\tau)e^{ix\tau}\,d\tau \ne 0, \qquad -\infty<x<\infty . \tag{1} \]
Korenblum \((^3)\) studied the case \(\alpha(\tau)=c\tau\). He found that, instead of (1), the necessary and sufficient conditions** are
\[ \tilde f(z)\ne 0,\qquad |\operatorname{Im} z|\leq c,\qquad -\infty<\operatorname{Re} z<\infty, \]
\[ \varlimsup_{x\to\infty}\ln|\tilde f(x)|/e^{\pi x/2c}=0,\qquad \varlimsup_{x\to-\infty}\ln|\tilde f(x)|/e^{-\pi x/2c}=0. \tag{2} \]
The question naturally arises: for which \(\alpha(\tau)\) is condition (1) sufficient for the completeness of the system \(\{f_x(\tau)\}\) in \(L_\alpha\). Let \(N\) be the class of such functions \(\alpha(\tau)\). In the present paper a sufficient condition is obtained for \(\alpha(\tau)\) to belong to the class \(N\), which is necessary if \(\alpha(\tau)=o(\tau)\) and \(\tau\alpha'(\tau)/\alpha(\tau)\) is nonincreasing. It is shown that this condition is at the same time, in an analogous way, necessary and sufficient for the regularity of \(L_\alpha\), regarded as a normed ring with multiplication given by convolution. For nonregular rings \(L\), necessary completeness conditions are obtained which are analogous to conditions (2).
2°. Theorem 1. If
\[ \int_0^\infty \frac{\alpha(\tau)\,d\tau}{1+\tau^2}<\infty, \tag{3} \]
then the ring \(L_\alpha\) is regular and the system \(\{f_x(\tau)\}\) is complete in \(L_\alpha\) if and only if (1) holds.
Proof. We first show the regularity of \(L_\alpha\). From the description of the maximal ideals of \(L_\alpha\) obtained in \((^4)\) it follows that it suffices, for arbitrary \(x_0\) and \(\varepsilon>0\), to construct a function \(\varphi(x)\in L_\alpha\) such that \(\tilde\varphi(x_0)\ne 0\) and \(\tilde\varphi(x)=0\) for \(|x-x_0|\geq \varepsilon\). We may assume that \(x_0=0\) and
\[ \alpha(x)\leq \int_1^x \frac{\alpha(\tau)\,d\tau}{\tau} \]
* A system \(\{x_\nu\}\) of elements of a Banach space \(X\) is called complete in \(X\) if the closure of the linear span of the system coincides with \(X\).
** In what follows, \(\tilde f(x)\) denotes the Fourier transform of the function \(f(x)\), defined by the equality in formula (1).
for \(x \geqslant 2\). Put
\[ \beta(\tau)=\alpha(\tau)\left(\int_{\tau}^{\infty}\frac{\alpha(\theta)\,d\theta}{1+\theta^2}\right)^{-1/2},\qquad \tau \geqslant 0,\qquad \beta(-\tau)=\beta(\tau). \]
Using the condition imposed on \(\alpha(\tau)\), we obtain, for \(A>1\),
\[ \int_{0}^{\infty}\frac{\beta(\tau)\,d\tau}{1+\tau^2}<\infty;\qquad \beta(A\tau)\leqslant A^2\beta(\tau);\qquad \lim_{x\to\infty}\left(\int_{1}^{x}\frac{\alpha(\tau)\,d\tau}{\tau} -\int_{1}^{x/A}\frac{\beta(\tau)\,d\tau}{\tau}\right)=-\infty . \tag{4} \]
Next, let \(\omega(\tau)\) be the function inverse to \(\beta(\tau)\), and for \(x \geqslant 1\)
\[ K(x)=\exp\left(\int_{1}^{x}\frac{\alpha(\tau)\,d\tau}{\tau}\right),\qquad N(x)=\exp\left(\int_{0}^{\omega(x)}\beta'(\tau)\ln \tau\,d\tau\right),\qquad F(x)=\min_{\tau\geqslant 1} N(\tau)x^{-\tau}. \]
Then \(N'(\beta(x))/N(\beta(x))=\ln x\), and therefore \(F(x)=N(\beta(x))x^{-\beta(x)}\), i.e.
\[ F(x)=\exp\left(-\int_{1}^{x}\frac{\beta(\tau)\,d\tau}{\tau}\right). \tag{5} \]
Moreover, for \(x \geqslant 1\) we have
\[ \frac{N'(x)}{N(x)}=\ln \omega(x),\qquad \ln\frac{N(x+1)}{N(x)}\geqslant \ln\omega(x),\qquad \frac{N(x)}{N(x+1)}\leqslant \int_{x-1}^{x}\frac{d\theta}{\omega(\theta)}, \]
hence, by virtue of (4),
\[ \sum_{n=2}^{\infty}\frac{N(n)}{N(n+1)} \leqslant \int_{0}^{\infty}\frac{d\theta}{\omega(\theta)} =O(1)+\int_{0}^{\infty}\frac{\beta(\tau)\,d\tau}{1+\tau^2}<\infty . \tag{6} \]
Now put
\[ \mu_2=\mu_3=1/N(2),\qquad \mu_j=N(j-3)/N(j-2), \]
\[ 4\leqslant j<\infty \quad\text{and}\quad \psi(x)=\prod_{n=2}^{\infty}\frac{\sin \mu_n x}{\mu_n x},\qquad \psi_\delta(x)=\psi(\delta x). \]
By (6), \(\psi(x)\) is an entire function of some finite degree. Since
\[
|\psi(x)|=O(1)\min_{j\geqslant 2} N(j-1)|x|^{-j},
\]
we have \(|\psi(x)|=O(F(|x|))\). Hence, from (4), (5) it follows that
\[ \int_{-\infty}^{\infty}|\psi_\delta(x)|e^{\alpha(x)}\,dx =O(1)\int_{1}^{\infty}F(\delta x)\exp\left(\int_{1}^{x}\frac{\alpha(\tau)\,d\tau}{\tau}\right)\,dx<\infty . \]
Thus \(\psi_\delta(x)\in L_\alpha\) for \(\delta>0\), and from the Wiener–Paley theorem we obtain that \(\widetilde{\psi}_\delta(x)=0\) for \(|x|\geqslant \delta\sigma\), while \(\widetilde{\psi}_\delta(0)=0\). Therefore the function \(\varphi(x)=\psi_{\varepsilon/\sigma}(x)\) is the desired one, and the regularity of \(L_\alpha\) is proved.
From the regularity of \(L_\alpha\) and the results of G. E. Shilov \({}^{(2)}\) it follows that, in order to prove the second assertion, it is enough to show the density in \(L_\alpha\) of functions with finite Fourier transforms. Let \(p(x)\in L_\alpha\). Put
\[ p_A(x)=Aq\int_{-\infty}^{\infty}p(x-t)\psi_A(t)\,dt,\qquad q=\left(\int_{-\infty}^{\infty}\psi(t)\,dt\right)^{-1}. \]
Obviously, \(p_A(x)\in L_\alpha\) and \(\widetilde{p}_A(x)\) is finite. We shall show that
\[ \lim_{A\to\infty}\|p-p_A\|=0. \]
For \(\mu>0\) and \(A\to\infty\) we have
\[ \int_{-\infty}^{\infty}\int_{|t|>\mu}|p(x)-p(x-t)|\,|\psi_A(t)|\,Aq\,dt\,e^{\alpha(x)}\,dx = O(1)\|p\|\int_{|t|>A\mu}|\psi(t)|e^{\alpha(t)}\,dt, \]
\[ \int_{-\infty}^{\infty}\int_{|t|<\mu}|p(x)-p(x-t)|\,|\psi_A(t)|\,Aq\,dt\,e^{\alpha(x)}\,dx\leqslant \]
\[ \leqslant \int_{|t|<\mu} |\psi_A(t)|\, A q \int_{-\infty}^{\infty} \bigl|p(x)e^{\alpha(x)} - p(x-t)e^{\alpha(x-t)}\bigr|\, dx\, dt + \]
\[ + \max_{|x|<\infty,\ |t|<\mu} \bigl|e^{\alpha(x)-\alpha(x-t)} - 1\bigr| \int_{|t|<\mu} |\psi_A(t)|\, A q\, dt = o(1) + O(\mu). \]
Consequently,
\[ \overline{\lim}_{A\to\infty} \|p-p_A\| = O(\mu). \]
Letting \(\mu \to 0\), we obtain
\[ \lim_{A\to\infty} \|p-p_A\| = 0, \]
as was required.
\(3^\circ\). Theorem 2. Let \(\alpha(\tau)=o(\tau)\), let \(\tau \alpha'(\tau)/\alpha(\tau)\) be nondecreasing,
\[ \int_0^\infty \frac{\alpha(\tau)\, d\tau}{1+\tau^2}=\infty. \]
Then the Fourier transforms of functions from \(L_\alpha\) form a quasi-analytic class of functions, and the ring \(L_\alpha\) is nonregular.
Theorem 3. If \(\alpha(\tau)\) satisfies the conditions of Theorem 2 and the system \(\{f_x(\tau)\}\) is complete in \(L_\alpha\), then (1) holds and, for real \(A\),
\[ \overline{\lim}_{x\to\infty} \frac{\ln |\widetilde f(x)|}{\xi\bigl(\tfrac12\pi(x-A)\bigr)} = 0, \qquad \overline{\lim}_{x\to-\infty} \frac{\ln |\widetilde f(x)|}{\xi\bigl(\tfrac12\pi(x+A)\bigr)} = 0, \tag{7} \]
where \(\xi(x)\) is the inverse function of the function
\[ \eta(x)=\int_0^x \frac{\alpha(\tau)\, d\tau}{1+\tau^2}. \]
Let us note that (7) coincides with (2) when \(\alpha(x)=cx\). Denote by \(I_{\gamma,A}\) the set of functions \(f\in L_\alpha\) for which
\[ \overline{\lim}_{x\to\infty} \frac{\ln |\widehat f(x)|}{\xi\bigl(\tfrac12\pi(x-A)\bigr)} \leqslant \gamma < 0. \]
Theorem 3 follows from the following assertion, which is of independent interest.
Theorem 4. Let \(\alpha(\tau)\) satisfy the conditions of Theorem 2. Then for arbitrary \(\gamma\) and \(A\) the set \(I_{\gamma,A}\) contains a function whose Fourier transform does not vanish, and there exists a function \(g(t)\), for which
\[ 0< \operatorname{vrai\,sup}_{-\infty<\tau<\infty} \bigl|g(\tau)e^{-\alpha(\tau)}\bigr| < \infty, \qquad \int_{-\infty}^{\infty} g(x-t) f(t)\, dt \equiv 0 \tag{8} \]
for all \(f\in I_{\gamma,A}\).
We shall need the following lemmas.
Lemma 1. Let \(\omega(t)\) be the inverse function for the function \(2\alpha(t)/\pi\),
\[ z=x+iy,\qquad \Phi(z)=\prod_{n=1}^{\infty}\left(1+\frac{z}{\omega(n)}\right)e^{-z/\omega(n)}. \]
Then we have
\[ \frac{c_1}{1+|y|^{c_2}} \leqslant \Phi(iy)e^{-\alpha(y)} \leqslant \Phi_1(y), \qquad \Phi_1(y)=\Phi_1(-y); \]
\[ \Phi_1(y)\uparrow,\qquad \int_0^\infty \frac{\ln \Phi_1(y)}{1+y^2}\, dy < \infty, \tag{9} \]
and, for \(x\geqslant 0\),
\[ |\Phi(x)| \leqslant c_4 e^{c_3 x} \exp\left( -\frac{2x}{\pi}\int_0^x \frac{\alpha(\tau)\, d\tau}{1+\tau^2} \right), \tag{10} \]
\[ |\Phi(z)| \geqslant \frac{c_5}{1+|y|^{c_6}} e^{-c_7 x} \exp\left( -\frac{2x}{\pi}\int_0^x \frac{\alpha(\tau)\, d\tau}{1+\tau^2} \right), \tag{11} \]
where \(c_i\) are positive constants.
\[ \text{* } 1/g(t)\ \text{depends on } \gamma \text{ and } A. \]
Lemma 2. If \(f\in I_{\gamma,A}\), then for \(x\geqslant 0\)
\[ \int_0^\infty |\widetilde f(p)|e^{xp}\,dp \leqslant c_9\exp\left(\frac{2x}{\pi}\int_0^x\frac{\alpha(\tau)\,d\tau}{1+\tau^2}+c_8x\right), \]
where \(c_8\) depends only on \(\gamma,A\).
Proof of Theorem 3. Put
\[ R(y)=\left(\frac{\sin y}{y}\right)^B\frac{1}{\Phi(iy)}. \]
By virtue of 9, for sufficiently large \(B\), \(R(y)\in L_\alpha\), and from (11) it follows that, for \(x>0,\ \sigma>0\),
\[ |\widetilde R(x)|\leqslant \exp\left(-\sigma x+c_{10}\sigma+\frac{2\sigma}{\pi}\int_0^\sigma \frac{\alpha(\tau)\,d\tau}{1+\tau^2}\right). \]
Taking \(\sigma=\xi(1/2\pi(x-H))\), for large \(H\) we obtain
\[ |\overline R(x)|\leqslant \exp\left(-\frac12 H\,\xi(1/2\pi(x-H))\right). \]
It is not difficult to show that, for any \(\delta>0\),
\[ \lim_{x=\infty}\frac{\xi(x-\delta)}{\xi(x)}=0. \]
Therefore one can choose \(\eta\) so that
\[ \widetilde H(y)=e^{i\eta y}R(y)\in I_{\gamma,A}, \]
and \(\widetilde H(y)\) does not vanish. The first assertion of the theorem is proved.
Next, put
\[ \mu(z)=\frac{2}{\pi}\int_{-\infty}^{\infty} \frac{\ln\Phi_1(\rho)\,d\rho}{i\rho-(z+1)}. \]
Then \(\mu(z)\) is holomorphic for \(\operatorname{Re}z>-1\), and by virtue of (9)
\[ \operatorname{Re}\mu(z)\leqslant -\frac{z\ln\Phi_1(z)}{\pi} \int_{\rho>|z|}\frac{(x+1)\,d\rho}{(\rho-y)^2+(x+1)^2}. \]
The function \(s(z)=e^{\mu(z)}\) is holomorphic and bounded for \(\operatorname{Re}z\geqslant0\), and
\[ |s(iy)|\leqslant \frac{1}{\Phi_1(|y|)}. \]
Put
\[ f\in I_{\gamma,A} \quad\text{and}\quad f_1(x)=\int_{-\infty}^{\infty}e^{-(x-\tau)^2}f(\tau)\,d\tau. \]
In this case
\[ \max_{-\infty<x<\infty}|f_1(x)e^{\alpha(x)}|<\infty,\qquad f_1(x)\in I_{\gamma,A}, \]
and, by Lemma 2, for \(x\geqslant0\),
\[ |f_1(iz)|=O(1)\left|\int_{-\infty}^{\infty}\widetilde f_1(p)e^{zp}\,dp\right| =O(1)\exp\left(\frac{2x}{\pi}\int_0^x \frac{\alpha(\tau)\,d\tau}{1+\tau^2}+c_8x\right). \]
Hence, and from (9), it follows that for any real \(\beta\) the function
\[ Q_\beta(z)=\Phi(z)s(z)e^{-2c_8z}f_1(iz+\beta) \]
is holomorphic in the right half-plane, bounded on the axis \(y\) and on the positive semiaxis \(x\), continuous for \(x\geqslant0\), and of order \(\rho<2\). Applying the Phragmén–Lindelöf principle, we find that \(Q_\beta(z)\) is bounded in the right half-plane. Then, by Cauchy’s theorem,
\[ \int_{-\infty}^{\infty}\frac{Q_\beta(iy)\,dy}{(1+iy)^2}=0. \]
Consequently, for \(-\infty<\beta<\infty\),
\[ \int_{-\infty}^{\infty}\Phi(iy)s(iy)e^{-2c_8iy}f_1(\beta-y)\frac{dy}{(1-iy)^2} = \int_{-\infty}^{\infty}e^{-(\beta-\tau)^2} \int_{-\infty}^{\infty} f(\tau-v)g(v)\,dv\,d\tau=0, \tag{12} \]
where
\[ g(v)=\Phi(iv)s(iv)e^{-2ic_8v}(1+iv)^{-2}. \]
From (12), by the usual device, we derive
\[ \int_{-\infty}^{\infty} f(\tau-v)g(v)\,dv=0. \]
Since \(g\) does not depend on \(f\) and, by virtue of (10),
\[ 0< \max_{-\infty<v<\infty}|g(v)e^{-\alpha(v)}|<\infty, \]
the last equality is equivalent to the last assertion of Theorem 3.
Leningrad Civil Engineering Institute
Received
18 V 1966
References
- N. Wiener, Ann. Math., 33, No. 1 (1932).
- G. E. Shilov, On regular normed rings, Moscow–Leningrad, 1947.
- B. I. Korenblum, Tr. Moscow Math. Soc., 7, 121 (1958).
- I. M. Gelfand, D. A. Raikov, G. E. Shilov, Uspekhi Mat. Nauk, 2 (12) (1946).