Full Text
D. L. BERMAN
POLYNOMIAL OPERATIONS COMMUTATIVE WITH SHIFT
(Presented by Academician V. I. Smirnov on 6 January 1962)
- Let us denote by \(\widetilde C\) the space of all continuous \(2\pi\)-periodic functions \(f(x)\) with norm
\[ \|f\|=\max_{0\le x\le 2\pi}|f(x)|; \]
by \(\Pi_n\) the set of all trigonometric polynomials of order \(\le n\); by \(M_n\) the set of all linear operations from \(\widetilde C\) into \(\widetilde C\) mapping \(\widetilde C\) into \(\Pi_n\); and by \(N_n\) the subset of \(M_n\) consisting of operators commutative with shift. Thus, if \(U_n\in N_n\), then for any \(f\in\widetilde C\) and for any \(-\infty<t<\infty\) the equality \(U(f_t)=[U_n(f)]_t\) holds, where \(f_t(x)=f(x+t)\).
Let \(U_n\in M_n\). Introduce the operator
\[
\widetilde U_n(f,x)=\frac{1}{2\pi}\int_0^{2\pi} U_n(f_t,x-t)\,dt .
\tag{1}
\]
It is obvious that \(U_n\in M_n\).
\(\widetilde U_n\) has the following properties: 1) \(\widetilde U_n\in N_n\); 2) if \(U_n\in N_n\), then \(\widetilde U_n=U_n\); this means that for any \(f\in\widetilde C\) the equality \(\widetilde U_n(f)=U_n(f)\) holds; 3) \(\|\widetilde U_n\|\le \|U_n\|\). Properties 1) and 2) are obvious. Property 3) is proved in \((1)\). We give examples.
Let \(U_n(f)=L_n(f)\), where \(L_n(f)\) is the Lagrange interpolation polynomial of order \(n\), constructed for an arbitrary system of interpolation nodes \(0\le x_0<x_1<\cdots<x_{2n}<2\pi\); then \({}^{(2)}\) \(\widetilde U_n(f)=S_n(f)\), where \(S_n(f)\) is the partial sum of order \(n\) of the Fourier series of \(f(x)\).
Let \(U_n(f)=\lambda_n(f)\), where
\[
\lambda_n(f,x)=\int_0^{2\pi} f(t)K_{n,p}(x,t)\,dt,
\tag{2}
\]
where \(K_{n,p}(x,t)\) is a trigonometric polynomial of order \(n\) with respect to \(x\) and of order \(p\) with respect to \(t\), \(p\le n\); then
\[
\widetilde U_n(f,x)=\int_0^{2\pi} f(z)\,dz\cdot \frac{1}{2\pi}\int_0^{2\pi} K_{n,p}(x-t,z-t)\,dt .
\tag{3}
\]
Let
\[
U_n(f,x)=\frac{2\pi}{2n+1}\sum_{j=0}^{2n} f\!\left(\frac{2\pi j}{2n+1}\right)K_{n,p}\!\left(x,\frac{2\pi j}{2n+1}\right),
\tag{4}
\]
where \(K_{n,p}(x,t)\) has the properties indicated earlier; then \(\widetilde U_n\) is again found according to (3).
- Let \(\widetilde U_n^{\langle i\rangle}\in M_n\), \(i=1,2\). Suppose that
\[ U_n^{\langle 1\rangle}(f)=U_n^{\langle 2\rangle}(f),\qquad f\in\Pi_n . \tag{5} \]
It does not at all follow from equality (5) that \(U_n^{\langle 1\rangle}=U_n^{\langle 2\rangle}\). However, the following theorem is true:
Theorem 1. If equality (5) holds on \(\Pi_n\), then \(\widetilde U_n^{\langle 1\rangle}=\widetilde U_n^{\langle 2\rangle}\). If \(U_n^{\langle 2\rangle}\in N_n\), then \(\widetilde U_n^{\langle 1\rangle}=U_n^{\langle 2\rangle}\).
Proof. By Lemma 2, from (3),
\[ \widetilde U_n^{\langle i\rangle}(f)=\widetilde U_n^{\langle i\rangle}(S_n(f)),\quad i=1,2. \tag{6} \]
Since \(S_n(f)\in\Pi_n\), by the hypothesis of the theorem,
\[ U_n^{\langle 1\rangle}(S_n(f))=U_n^{\langle 2\rangle}(S_n(f)). \tag{7} \]
The required assertion follows from (6) and (7). The second part of the theorem follows from the first part and from property 2) of the operator \(\widetilde U_n\).
Special cases of Theorem 1 are known \((^4)\). Let \(U_n\in M_n\) and \(U_n(f)=\lambda_n(f)\), \(f\in\Pi_n\), where \(\lambda_n(f)\) is defined according to (2). Then, by Theorem 1, for any \(f\in\widetilde C\),
\[ \frac{1}{2\pi}\int_0^{2\pi} U_n(f_t,x-t)\,dt = \int_0^{2\pi} f(z)\,dz\int_0^{2\pi} K_{n,p}(x-t,z-t)\,dt. \]
Theorem 2. Suppose that (5) holds on \(\Pi_n\), where \(U_n^{\langle i\rangle}\in M_n,\ i=1,2\). Then
\[ \|U_n^{\langle 1\rangle}\|\geq \|\widetilde U_n^{\langle 2\rangle}\|;\qquad \|U_n^{\langle 2\rangle}\|\geq \|\widetilde U_n^{\langle 1\rangle}\|. \tag{8} \]
Proof. By Theorem 1, \(\widetilde U_n^{\langle 1\rangle}=\widetilde U_n^{\langle 2\rangle}\). Therefore \(\|\widetilde U_n^{\langle 1\rangle}\|=\|\widetilde U_n^{\langle 2\rangle}\|\). By property 3) of the operator \(\widetilde U_n\), we have \(\|\widetilde U_n^{\langle i\rangle}\|\leq \|U_n^{\langle i\rangle}\|\). Consequently, \(\|U_n^{\langle 1\rangle}\|\geq \|\widetilde U_n^{\langle 2\rangle}\|\). The second inequality in (8) is proved in the same way.
It is useful to compare this theorem with the results of \((^{3,5-7})\).
In connection with Theorem 1 the question arises: in what case does the validity of equality (5) on \(\Pi_n\) imply its validity for any \(f\in\widetilde C\)? The answer to this question is given by Theorem 3.
Theorem 3. In order that the validity of equality (5) on \(\Pi_n\) imply its validity for any \(f\in C\), it is necessary and sufficient that \(U_n^{\langle 3\rangle}\in N_n\), where \(U_n^{\langle 3\rangle}=U_n^{\langle 2\rangle}-U_n^{\langle 1\rangle}\).
Proof. Suppose equality (5) holds for any \(f\in\widetilde C\). Then \(U_n^{\langle 3\rangle}(f)=0\). Therefore, for any \(-\infty<t<\infty\),
\[ U_n^{\langle 3\rangle}(f_t)=[U_n^{\langle 3\rangle}(f)]_t=0. \]
Conversely, suppose \(U_n^{\langle 3\rangle}\in N_n\). By property 2) of the operator \(\widetilde U_n\), \(U_n^{\langle 3\rangle}=\widetilde U_n^{\langle 3\rangle}\). By hypothesis, \(U_n^{\langle 1\rangle}(f)=U_n^{\langle 2\rangle}(f)\), \(f\in\Pi_n\); hence \(U_n^{\langle 3\rangle}(f)=0\), \(f\in\Pi_n\). But in this case, by Theorem 1, \(\widetilde U_n^{\langle 3\rangle}(f)=0\) for any \(f\in\widetilde C\); hence \(U_n^{\langle 3\rangle}=0\). Let \(U_n\in M_n\). The question arises: when does there exist an operator from \(N_n\) that coincides on \(\Pi_n\) with \(U_n\)? The answer is given by Theorem 4.
Theorem 4. In order that, for a given \(U_n\in M_n\), there exist an operator \(U_n^{\langle 0\rangle}\in N_n\) coinciding on \(\Pi_n\) with the operator \(U_n\), it is necessary and sufficient that the condition
\[ \widetilde U_n(f)=U_n(f),\quad f\in\Pi_n \tag{9} \]
be satisfied.
If this condition is satisfied, then there exists a unique operator \(\widetilde U_n\) satisfying the stated condition.
Proof. Suppose there exists an operator \(U_n^{\langle 0\rangle}\in N_n\) coinciding on \(\Pi_n\) with the operator \(U_n\). By Theorem 1, \(\widetilde U_n^{\langle 0\rangle}=\widetilde U_n\). By property 2) of the operator \(\widetilde U_n\), \(\widetilde U_n^{\langle 0\rangle}=U_n^{\langle 0\rangle}\). It follows that \(U_n^{\langle 0\rangle}=\widetilde U_n\).
Thus (9) has been established. From these arguments follows the uniqueness of the operator from \(N_n\) that coincides on \(\Pi_n\) with the given \(U_n\). Sufficiency is obvious, since \(\widetilde U_n \in N_n\).
Remark. With the help of Theorem 4 it is not difficult to see that if the kernel of the operator (2) satisfies the condition \(K_{n,n}(x,t) \ne K_n(x-t)\), then for the operator (2) one cannot construct an operator from \(N_n\) satisfying condition (5).
3. Let, for the operator \(U_n \in M_n\), there exist a convolution
\[ \sigma_n(f,x)=\int_0^{2\pi} f(x+t)\Phi_n(t)\,dt, \]
where \(\Phi_n(t)\) is a polynomial of order \(n\), such that on \(\Pi_n\) the equality
\[ U_n(f)=\sigma_n(f),\qquad f\in \Pi_n \tag{10} \]
holds. In this case, as is known \((^3)\), for any \(f\in \widetilde C\) the equality
\[ \widetilde U_n(f)=\sigma_n(f) \]
holds.
It turns out that requirement (10) is superfluous, for the following theorem is true.
Theorem 5. Let \(U_n\in M_n\); then for any \(f\in \widetilde C\) the equality
\[ \frac{1}{2\pi}\int_0^{2\pi} U_n(f_t,x-t)\,dt = \int_0^{2\pi} f(x+t)\Phi_n(t)\,dt, \tag{11} \]
holds, where \(\Phi_n(x)=\widetilde U_n[D_n(z)-x]\). \((D_n(z)\) is the Dirichlet kernel of order \(n\).)
Proof. According to (3) we have
\[ \widetilde U_n(f)=\widetilde U_n[S_n(f)]. \tag{12} \]
We compute the right-hand side
\[ \widetilde U_n[S_n(f),x] = \frac{1}{2\pi}\int_0^{2\pi} U_n[(S_n(f))_t,x-t]\,dt. \tag{13} \]
Since
\[ [S_n(f)]_t(z)=\int_0^{2\pi} f_t(t_1)D_n(z-t_1)\,dt_1, \]
it follows from (13) that
\[ \widetilde U_n[S_n(f),x] = \frac{1}{2\pi}\int_0^{2\pi}dt \int_0^{2\pi} f_t(t_1)\,U_n[D_n(z-t_1),x-t]\,dt_1. \]
Therefore
\[ \widetilde U_n[S_n(f),x] = \frac{1}{2\pi}\int_0^{2\pi}dt_1 \int_0^{2\pi} f(t+t_1)\widetilde U_n[D_n(z-t_1),x-t]\,dt = \]
\[ = \frac{1}{2\pi}\int_0^{2\pi}dt_1 \int_0^{2\pi} f(z_1)\widetilde U_n[D_n(z-t_1),x-z_1+t]\,dz_1. \tag{14} \]
Since \(\widetilde U_n\in N_n\), we have
\[ \widetilde U_n[D_n(z-t_1),x-z_1+t_1] = \widetilde U_n[D_n(z),x-z_1]. \]
Thus, equality (14) takes the form
\[ \tilde U_n [S_n(f),\, x] = \int_0^{2\pi} f(z_1)\Phi_n(z_1-x)\,dz_1, \tag{15} \]
where
\[ \Phi_n(x)=\tilde U_n[D_n(z),\, -x]. \]
From equalities (12) and (15) the following theorem follows:
Theorem 6. For any \(U_n\in M_n\) the inequality
\[ \|U_n\|\ge \int_0^{2\pi}\left|\tilde U_n[D_n(z),\, x]\right|\,dx \tag{16} \]
holds.
Proof. From Theorem 5 it follows that \(\|\tilde U_n\|\) is equal to the norm of the convolution on the right-hand side of (15), which is equal to the integral on the right-hand side of (15). To complete the proof it remains only to note that, by property 3) of the operator \(\tilde U_n\), \(\|\tilde U_n\|\le \|U_n\|\).
For simplicity we have considered the case of the space \(\tilde C\). The results of the note admit extension also to other functional spaces. They remain valid also in the case of functions defined on bicompact commutative groups.
Leningrad Institute of Soviet Trade
named after Fr. Engels
Received
4 I 1962
References
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