Abstract Generated abstract
The paper studies sharp norm estimates for differential operators acting on entire functions of exponential type at most ν that belong to L2 on the real axis. Using integral representations for such functions and their L2 conjugates, together with the Cauchy, Bunyakovsky inequality and Parseval identities, it derives an exact bound for infinite linear combinations of shifted derivatives of a function and its conjugate under a natural convergence condition on the coefficients. The authors identify an extremal function showing equality in the main inequality and compute the constant explicitly in several special cases, including combinations of values or derivatives at conjugate imaginary shifts. The resulting inequalities extend and sharpen earlier one dimensional estimates, with multidimensional analogues indicated.
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MATHEMATICS
I. I. IBRAGIMOV, A. S. JAFAROV
ESTIMATE OF A DIFFERENTIAL OPERATOR IN THE CLASS OF ENTIRE FUNCTIONS OF FINITE DEGREE
(Presented by Academician V. I. Smirnov on 20 VI 1963)
Denote by \(W_\nu^{(2)}\) the class of entire functions \(f(z)\) of degree not exceeding \(\nu\), for which the condition is satisfied:
\[
(\|f\|_2)^2=\int_{-\infty}^{\infty}|f(x)|^2\,dx<+\infty .
\]
For a function \(f(z)\) and the function \(\widetilde f(z)\) conjugate to it in the sense of \(\mathscr L_2\), by virtue of the formula for the convolution of two functions from \(\mathscr L_2\), the following identity holds (see \(\left({}^{1}\right)\)):
\[
f(z)=\frac1\pi\int_{-\infty}^{\infty} f(t)\frac{\sin\nu(t+z)}{t+z}\,dt,
\tag{1}
\]
\[
\widetilde f(z)=-\frac1\pi\int_{-\infty}^{\infty} f(t)\frac{2\sin^2\frac12\nu(t+z)}{t+z}\,dt.
\tag{2}
\]
In the present note an exact estimate is given for the expression *
\[
\Phi(x)=\sum_{k=0}^{\infty}\{a_k f^{(m_k)}(x+ic_k)+b_k\widetilde f^{(m_k)}(x+ic_k)\},
\tag{3}
\]
where \(f\in W_\nu^{(2)}\); \(a_k, b_k\) are arbitrary complex numbers; \(c_k\) are arbitrary real numbers; \(m_0\le m_1\le m_2\le\cdots\) are nonnegative integers such that the series
\[
\sum_{k=0}^{\infty}\nu^{m_k}(|a_k|+|b_k|)e^{\nu|c_k|}
\tag{4}
\]
converges.
On the basis of (1) and (2) we have
\[
\Phi(x)=\frac1\pi\int_{-\infty}^{\infty} f(t)\left\{\sum_{k=0}^{\infty}\left(a_k\frac{\partial^{m_k}}{\partial x^{m_k}}\left[\frac{\sin\nu(t+x+ic_k)}{t+x+ic_k}\right]-\right.\right.
\]
\[
\left.\left.
-b_k\frac{\partial^{m_k}}{\partial x^{m_k}}\left[\frac{2\sin^2\frac12\nu(t+x+ic_k)}{t+x+ic_k}\right]\right)\right\}\,dt .
\]
Applying Bunyakovsky’s inequality, we obtain
\[
|\Phi(x)|\le\frac1\pi B_\nu\|f\|_2,
\tag{5}
\]
* Some special cases of the result of this note were obtained in \(\left({}^{5}\right)\).
where
\[ B_\nu^2=\nu \int_{-\infty}^{\infty}\left|\sum_{k=0}^{\infty}\left(a_k\nu^{m_k}\left\{\frac{\sin(\tau+i\nu c_k)}{\tau+i\nu c_k}\right\}^{(m_k)} -2b_k\nu^{m_k}\left\{\frac{\sin^2 \tfrac12 \nu(\tau+i\nu c_k)}{\tau+i\nu c_k}\right\}^{(m_k)}\right)\right|^2\,d\tau, \tag{6} \]
\[ \{\varphi(\tau+i\alpha)\}^{(m_k)} = \left.\frac{d^{m_k}\varphi(z)}{dz^{m_k}}\right|_{z=\tau+i\alpha}. \]
We note that inequality (5) for \(x=0\) becomes an equality for the function
\[ f_0(z)=\sum_{k=0}^{\infty}\left\{ \bar a_k\,\frac{d^{m_k}}{dz^{m_k}}\left[\frac{\sin \nu(z-ic_k)}{z-ic_k}\right] -\bar b_k\,\frac{d^{m_k}}{dz^{m_k}}\left[\frac{2\sin^2 \tfrac12\nu(z-ic_k)}{z-ic_k}\right]\right\}, \]
which belongs to the class \(W_\nu^{(2)}\) under the condition of convergence of the series (4). This follows from the fact that if we have a sequence \(f_m(z)\in W_\nu^{(2)}\) such that \(\|f_m\|_2\le M<\infty\), and if the series \(\sum_{k=0}^{\infty} A_k\), where \(A_k\) are arbitrary numbers, converges absolutely, then
\[ f(z)=\sum_{k=0}^{\infty} A_k f_k(z) \tag{7} \]
is an entire function of degree not exceeding \(\nu\), and the series (7) converges uniformly on the whole real axis. This follows from the corresponding inequalities for entire functions of finite degree (see, for example, \((^2)\), inequality (2.7)) and from Weierstrass’ theorem on the uniform convergence of a series of analytic functions.
Let us consider various special cases. Suppose first that all \(c_k\) are equal to zero, and \(a_k=\alpha_k+i\beta_k,\ b_k=\gamma_k+id_k\), where \(\alpha_k,\ \beta_k,\ \gamma_k\), and \(d_k\) are real \((k=0,1,2,\ldots)\).
Then from (6) we have:
\[ B_\nu^2=\nu\,(I_{\alpha,\gamma}+I_{\beta,d}), \]
where
\[ \begin{aligned} I_{s,r} &=\int_{-\infty}^{\infty}\left(\sum_{k=0}^{\infty}s_k\nu^{m_k}\left\{\frac{\sin\tau}{\tau}\right\}^{(m_k)} -2r_k\nu^{m_k}\left\{\frac{\sin^2 \tfrac12\tau}{\tau}\right\}^{(m_k)}\right)^2\,d\tau \\ &=\sum_{i,j=0}^{\infty}s_i s_j\nu^{m_i+m_j} \int_{-\infty}^{\infty} \left(\frac{\sin\tau}{\tau}\right)^{(m_i)} \left(\frac{\sin\tau}{\tau}\right)^{(m_j)}\,d\tau \\ &\quad -4\sum_{i,j=0}^{\infty}s_i r_j\nu^{m_i+m_j} \int_{-\infty}^{\infty} \left(\frac{\sin\tau}{\tau}\right)^{(m_i)} \left(\frac{\sin^2 \tfrac12\tau}{\tau}\right)^{(m_j)}\,d\tau \\ &\quad +4\sum_{i,j=0}^{\infty}r_i r_j\nu^{m_i+m_j} \int_{-\infty}^{\infty} \left(\frac{\sin^2 \tfrac12\tau}{\tau}\right)^{(m_i)} \left(\frac{\sin^2 \tfrac12\tau}{\tau}\right)^{(m_j)}\,d\tau . \end{aligned} \tag{8} \]
We note that the Fourier transform in \(\mathcal L_2(-\infty,\infty)\) of the function
\[ h(x)= \begin{cases} \dfrac{\sqrt{2\pi}}{2}(ix)^k, & \text{if } |x|\le 1,\\[4pt] 0, & \text{if } |x|>1, \end{cases} \tag{9} \]
is \((\sin x/x)^{(k)}\). Denote by \(h_1(x)\) the function (9) for \(k=m_i\), and by \(h_2(x)\) the function (9) for \(k=m_j\). Then the Fourier transforms in \(\mathcal L_2(-\infty,\infty)\)
the functions \(h_1(x)\) and \(h_2(x)\) are respectively the functions \(\left(\dfrac{\sin x}{x}\right)^{(m_i)}\) and \(\left(\dfrac{\sin x}{x}\right)^{(m_j)}\).
On the basis of the generalized Parseval equality for functions from \(L_2(-\infty,\infty)\), we have:
\[ \int_{-\infty}^{\infty} \left(\frac{\sin x}{x}\right)^{(m_i)} \left(\frac{\sin x}{x}\right)^{(m_j)}\,dx = \int_{-1}^{1} \left(\frac{\sqrt{2\pi}}{2}\right)^2 (ix)^{m_i}(-ix)^{m_j}\,dx = \]
\[ = \begin{cases} (-1)^{m_j+(m_i+m_j)/2}\dfrac{\pi}{m_i+m_j+1}, & \text{if } m_i+m_j \text{ is even},\\[6pt] 0, & \text{if } m_i+m_j \text{ is odd}. \end{cases} \]
In an analogous way one can show that
\[ \int_{-\infty}^{\infty} \left(\frac{\sin t}{t}\right)^{(m_i)} \left(\frac{\sin^2 \tfrac12 t}{t}\right)^{(m_j)}\,dt = \]
\[ = \begin{cases} 0, & \text{if } m_i+m_j \text{ is even},\\[6pt] (-1)^{m_j+(m_i+m_j+1)/2}\dfrac{\pi}{2(m_i+m_j+1)}, & \text{if } m_i+m_j \text{ is odd}. \end{cases} \]
Moreover,
\[ \int_{-\infty}^{\infty} \left(\frac{\sin^2 \tfrac12 t}{t}\right)^{(m_i)} \left(\frac{\sin^2 \tfrac12 t}{t}\right)^{(m_j)}\,dt = \]
\[ = \begin{cases} (-1)^{m_j+(m_i+m_j)/2}\dfrac{\pi}{4(m_i+m_j+1)}, & \text{if } m_i+m_j \text{ is even},\\[6pt] 0, & \text{if } m_i+m_j \text{ is odd}. \end{cases} \]
Thus, from (8) we have:
\[ \begin{aligned} I_{s,r} &= \pi \sum_{i,j=0}^{\infty}{}' (-1)^{m_j+(m_i+m_j)/2} \frac{\nu^{m_i+m_j}}{m_i+m_j+1}\,s_i s_j \\ &\quad +2\pi \sum_{i,j=0}^{\infty}{}'' (-1)^{m_j+(m_i+m_j+1)/2} \frac{\nu^{m_i+m_j}}{m_i+m_j+1}\,s_i r_j \\ &\quad +\pi \sum_{i,j=0}^{\infty}{}' (-1)^{m_j+(m_i+m_j)/2} \frac{\nu^{m_i+m_j}}{m_i+m_j+1}\,r_i r_j, \tag{10} \end{aligned} \]
where the prime in the first and third sums means that these sums extend over all possible \(i,j=0,1,2,\ldots\) for which \(m_i+m_j\) is nonnegative even, and the double prime in the second sum means that it extends over all possible \(i,j=0,1,2,\ldots\) for which \(m_i+m_j\) is odd.
Consequently, in the case under consideration \(B_\nu\), entering into inequality (5), is equal to \(\{\pi\nu(I_{\alpha,\gamma}+I_{\beta,d})\}^{1/2}\), where \(I_{\alpha,\gamma}\) and \(I_{\beta,d}\) are computed by formula (10).
Let now \(m_0=m_1=k\), \(a_0=e^{-i\omega}\), \(a_1=e^{i\omega}\) (\(\omega\) is a real number), \(a_i=0\) \((i=2,3,\ldots)\), \(b_i=0\) \((i=0,1,2,\ldots)\), \(c_0=-c_1=c\). In this case, as is seen from (6):
\[ B_\nu^2 = \nu^{2k+1} \int_{-\infty}^{\infty} \left| e^{-i\omega} \left\{\frac{\sin(\tau+i\nu c)}{\tau+i\nu c}\right\}^{(k)} + e^{i\omega} \left\{\frac{\sin(\tau-i\nu c)}{\tau-i\nu c}\right\}^{(k)} \right|^2\,d\tau . \]
Let us note that, analogously to what was done above, we have:
\[ \int_{-\infty}^{\infty} \left\{\frac{\sin(x+ia)}{x+ia}\right\}^{(k)} \left\{\frac{\sin(x+ib)}{x+ib}\right\}^{(k)}\,dx = \frac{\pi}{2} \int_{-1}^{+1} x^{2k}e^{-x(a-b)}\,dx . \tag{11} \]
Thanks to this we find
\[ B_\nu^2=2\pi\nu^{2k+1}\left[\int_0^1 \tau^{2k}\operatorname{ch}2\nu c\tau\,d\tau+\frac{\cos 2\omega}{2k+1}\right]. \]
Thus, from (5) we obtain
\[ \begin{aligned} &\left|e^{-i\omega}f^{(k)}(x+ic)+e^{i\omega}f^{(k)}(x-ic)\right|\leq \\ &\leq \nu^k\left(\frac{2\nu}{\pi}\right)^{1/2} \left[\int_0^1 \tau^{2k}\operatorname{ch}2\nu c\tau\,d\tau+\frac{\cos 2\omega}{2k+1}\right]^{1/2} \|f\|_2 . \end{aligned} \tag{12} \]
In particular, for \(k=0\), from (13) it follows that
\[ \left|e^{-i\omega}f(x+ic)+e^{i\omega}f(x-ic)\right| \leq \left(\frac{2\nu}{\pi}\right)^{1/2} \left[\frac{\operatorname{sh}2\nu c}{2\nu c}+\cos 2\omega\right]^{1/2} \|f\|_2 . \tag{13} \]
Inequality (13) is an analogue of the corresponding inequality of R. P. Boas \((^3)\) and of inequality (5.1) from the work \((^4)\).
Suppose now that \(m_0=m_1=k\), \(a_0=e^{-i\omega}\), \(b_1=e^{i\omega}\), \(a_i=0\) \((i=1,2,\ldots)\), \(b_0=0\), \(b_i=0\) \((i=2,3,\ldots)\), \(c_0=-c_1=c\).
In this case, as is seen from (6),
\[ B_\nu^2=\nu^{2k+1}\int_{-\infty}^{\infty} \left| e^{-i\omega}\left\{\frac{\sin(t+i\nu c)}{t+i\nu c}\right\}^{(k)} - e^{i\omega}\left\{\frac{2\sin^2\frac12(t-i\nu c)}{t-i\nu c}\right\}^{(k)} \right|^2dt . \]
Then, on the basis of (11) and since
\[ \int_{-\infty}^{\infty} \left\{\frac{\sin(x+i\nu c)}{x+i\nu c}\right\}^{(k)} \left\{\frac{\sin^2\frac12(x+i\nu c)}{x+i\nu c}\right\}^{(k)} dx=0, \]
\[ \int_{-\infty}^{\infty} \left\{\frac{2\sin^2\frac12(x-i\nu c)}{x-i\nu c}\right\}^{(k)} \left\{\frac{2\sin\frac12(x+i\nu c)}{x+i\nu c}\right\}^{(k)} dx= \frac{\pi}{2}\int_{-1}^{+1} t^{2k}e^{2\nu ct}\,dt, \]
we obtain that
\[ B_\nu^2=2\pi\nu^{2k+1}\int_0^1 t^{2k}\operatorname{ch}2\nu ct\,dt . \]
Thus, from (5) it follows that
\[ \left|e^{-i\omega}f^{(k)}(x+ic)+e^{i\omega}\widetilde f^{(k)}(x-ic)\right| \leq \nu^k\left(\frac{2\nu}{\pi}\right)^{1/2} \left[\int_0^1 \tau^{2k}\operatorname{ch}2\nu c\tau\,d\tau\right]^{1/2} \|f\|_2 . \tag{14} \]
In particular, for \(k=0\), from (14) it follows that
\[ \left|e^{-i\omega}f(x+ic)+e^{i\omega}\widetilde f(x-ic)\right| \leq \left(\frac{2\nu}{\pi}\right)^{1/2} \left(\frac{\operatorname{sh}2\nu c}{2\nu c}\right)^{1/2} \|f\|_2 . \tag{15} \]
We note that inequalities (12)—(15) are sharp in the class \(W_\nu^{(2)}\), and the extremal functions are obtained from the expression \(f_0(z)\) by a corresponding choice of constants.
Multidimensional cases can be considered in an analogous manner.
Institute of Mathematics and Mechanics
Academy of Sciences of the Azerbaijan SSR
Received
1 VI 1962
CITED LITERATURE
- I. I. Ibragimov, Extremal properties of entire functions of finite degree, Baku, 1962.
- I. I. Ibragimov, Izv. AN SSSR, Ser. Mat., 23, 80 (1959).
- R. P. Boas, Math. Scand., 4, 29 (1956).
- I. I. Ibragimov, Izv. AN SSSR, Ser. Mat., 24, 605 (1960).
- I. I. Ibragimov, A. S. Dzhafarov, DAN, 138, No. 4, 751 (1961).