Silverman–Toeplitz theorem

In mathematics, the Silverman–Toeplitz theorem, first proved by Otto Toeplitz, is a result in series summability theory characterizing matrix summability methods that are regular. A regular matrix summability method is a linear sequence transformation that preserves the limits of convergent sequences. The linear sequence transformation can be applied to the divergent sequences of partial sums of divergent series to give those series generalized sums.

An infinite matrix ( a i , j ) i , j ∈ N {\displaystyle (a_{i,j})_{i,j\in \mathbb {N} }} with complex-valued entries defines a regular matrix summability method if and only if it satisfies all of the following properties:

lim i → ∞ a i , j = 0 j ∈ N (Every column sequence converges to 0.) lim i → ∞ ∑ j = 0 ∞ a i , j = 1 (The row sums converge to 1.) sup i ∑ j = 0 ∞ | a i , j | < ∞ (The absolute row sums are bounded.) {\displaystyle {\begin{aligned}&\lim _{i\to \infty }a_{i,j}=0\quad j\in \mathbb {N} &&{\text{(Every column sequence converges to 0.)}}\\[3pt]&\lim _{i\to \infty }\sum _{j=0}^{\infty }a_{i,j}=1&&{\text{(The row sums converge to 1.)}}\\[3pt]&\sup _{i}\sum _{j=0}^{\infty }\vert a_{i,j}\vert <\infty &&{\text{(The absolute row sums are bounded.)}}\end{aligned}}}

An example is Cesàro summation, a matrix summability method with

a m n = { 1 m n ≤ m 0 n > m = ( 1 0 0 0 0 ⋯ 1 2 1 2 0 0 0 ⋯ 1 3 1 3 1 3 0 0 ⋯ 1 4 1 4 1 4 1 4 0 ⋯ 1 5 1 5 1 5 1 5 1 5 ⋯ ⋮ ⋮ ⋮ ⋮ ⋮ ⋱ ) . {\displaystyle a_{mn}={\begin{cases}{\frac {1}{m}}&n\leq m\\0&n>m\end{cases}}={\begin{pmatrix}1&0&0&0&0&\cdots \\{\frac {1}{2}}&{\frac {1}{2}}&0&0&0&\cdots \\{\frac {1}{3}}&{\frac {1}{3}}&{\frac {1}{3}}&0&0&\cdots \\{\frac {1}{4}}&{\frac {1}{4}}&{\frac {1}{4}}&{\frac {1}{4}}&0&\cdots \\{\frac {1}{5}}&{\frac {1}{5}}&{\frac {1}{5}}&{\frac {1}{5}}&{\frac {1}{5}}&\cdots \\\vdots &\vdots &\vdots &\vdots &\vdots &\ddots \\\end{pmatrix}}.}

Formal statement

Let the aforementioned inifinite matrix ( a i , j ) i , j ∈ N {\displaystyle (a_{i,j})_{i,j\in \mathbb {N} }} of complex elements satisfy the following conditions:

lim i → ∞ a i , j = 0 {\displaystyle \lim _{i\to \infty }a_{i,j}=0} for every fixed j ∈ N {\displaystyle j\in \mathbb {N} }.
sup i ∈ N ∑ j = 1 i | a i , j | < ∞ {\displaystyle \sup _{i\in \mathbb {N} }\sum _{j=1}^{i}\vert a_{i,j}\vert <\infty };

and z n {\displaystyle z_{n}} be a sequence of complex numbers that converges to lim n → ∞ z n = z ∞ {\displaystyle \lim _{n\to \infty }z_{n}=z_{\infty }}. We denote S n {\displaystyle S_{n}} as the weighted sum sequence: S n = ∑ m = 1 n a n , m z m {\displaystyle S_{n}=\sum _{m=1}^{n}a_{n,m}z_{m}}.

Then the following results hold:

If lim n → ∞ z n = z ∞ = 0 {\displaystyle \lim _{n\to \infty }z_{n}=z_{\infty }=0}, then lim n → ∞ S n = 0 {\displaystyle \lim _{n\to \infty }{S_{n}}=0}.
If lim n → ∞ z n = z ∞ ≠ 0 {\displaystyle \lim _{n\to \infty }z_{n}=z_{\infty }\neq 0} and lim i → ∞ ∑ j = 1 i a i , j = 1 {\displaystyle \lim _{i\to \infty }\sum _{j=1}^{i}a_{i,j}=1}, then lim n → ∞ S n = z ∞ {\displaystyle \lim _{n\to \infty }{S_{n}}=z_{\infty }}.

Proof

Proving 1.

For the fixed j ∈ N {\displaystyle j\in \mathbb {N} } the complex sequences z n {\displaystyle z_{n}}, S n {\displaystyle S_{n}} and a i , j {\displaystyle a_{i,j}} approach zero if and only if the real-values sequences | z n | {\displaystyle \left|z_{n}\right|}, | S n | {\displaystyle \left|S_{n}\right|} and | a i , j | {\displaystyle \left|a_{i,j}\right|} approach zero respectively. We also introduce M = 1 + sup i ∈ N ∑ j = 1 i | a i , j | > 0 {\displaystyle M=1+\sup _{i\in \mathbb {N} }\sum _{j=1}^{i}\vert a_{i,j}\vert >0}.

Since | z n | → 0 {\displaystyle \left|z_{n}\right|\to 0}, for prematurely chosen ε > 0 {\displaystyle \varepsilon >0} there exists N ε ∈ N {\displaystyle N_{\varepsilon }\in \mathbb {N} }, so for every n > N ε {\displaystyle n>N_{\varepsilon }} we have | z n | < ε 2 M {\displaystyle \left|z_{n}\right|<{\frac {\varepsilon }{2M}}}. Next, for some N a = N a ( ε ) > N ε {\displaystyle N_{a}=N_{a}\left(\varepsilon \right)>N_{\varepsilon }} it's true, that ∑ m = 1 n | a n , m | < ε 2 ( max m ≤ N ε | z m | + 1 ) {\displaystyle \sum _{m=1}^{n}|a_{n,m}|<{\frac {\varepsilon }{2\left(\max _{m\leq N_{\varepsilon }}|z_{m}|+1\right)}}} for every n > N a ( ε ) {\displaystyle n>N_{a}\left(\varepsilon \right)}. Therefore, for every n > N a ( ε ) {\displaystyle n>N_{a}\left(\varepsilon \right)}

| S n | = | ∑ m = 1 n ( a n , m z m ) | ⩽ ∑ m = 1 n ( | a n , m | ⋅ | z m | ) = ∑ m = 1 N ε ( | a n , m | ⋅ | z m | ) + ∑ m = N ε + 1 n ( | a n , m | ⋅ | z m | ) < < max 1 ≤ m ≤ N ε ( | z m | ) ⋅ ∑ m = 1 N ε | a n , m | + ε 2 M ∑ m = N ε + 1 n | a n , m | ⩽ ε 2 + ε 2 M ∑ m = 1 n | a n , m | ⩽ ε 2 + ε 2 M ⋅ M = ε {\displaystyle {\begin{aligned}&\left|S_{n}\right|=\left|\sum _{m=1}^{n}\left(a_{n,m}z_{m}\right)\right|\leqslant \sum _{m=1}^{n}\left(\left|a_{n,m}\right|\cdot \left|z_{m}\right|\right)=\sum _{m=1}^{N_{\varepsilon }}\left(\left|a_{n,m}\right|\cdot \left|z_{m}\right|\right)+\sum _{m=N_{\varepsilon }+1}^{n}\left(\left|a_{n,m}\right|\cdot \left|z_{m}\right|\right)<\\&<\max _{1\leq m\leq N_{\varepsilon }}(|z_{m}|)\cdot \sum _{m=1}^{N_{\varepsilon }}|a_{n,m}|+{\frac {\varepsilon }{2M}}\sum _{m=N_{\varepsilon }+1}^{n}\left|a_{n,m}\right|\leqslant {\frac {\varepsilon }{2}}+{\frac {\varepsilon }{2M}}\sum _{m=1}^{n}\left|a_{n,m}\right|\leqslant {\frac {\varepsilon }{2}}+{\frac {\varepsilon }{2M}}\cdot M=\varepsilon \end{aligned}}}

which means, that both sequences | S n | {\displaystyle \left|S_{n}\right|} and S n {\displaystyle S_{n}} converge zero.

Proving 2.

lim n → ∞ ( z m − z ∞ ) = 0 {\displaystyle \lim _{n\to \infty }\left(z_{m}-z_{\infty }\right)=0}. Applying the already proven statement yields . Finally,lim n → ∞ ∑ m = 1 n ( a n , m ( z m − z ∞ ) ) = 0 {\displaystyle \lim _{n\to \infty }\sum _{m=1}^{n}{\big (}a_{n,m}\left(z_{m}-z_{\infty }\right){\big )}=0}

lim n → ∞ S n = lim n → ∞ ∑ m = 1 n ( a n , m z m ) = lim n → ∞ ∑ m = 1 n ( a n , m ( z m − z ∞ ) ) + z ∞ lim n → ∞ ∑ m = 1 n ( a n , m ) = 0 + z ∞ ⋅ 1 = z ∞ {\displaystyle \lim _{n\to \infty }S_{n}=\lim _{n\to \infty }\sum _{m=1}^{n}{\big (}a_{n,m}z_{m}{\big )}=\lim _{n\to \infty }\sum _{m=1}^{n}{\big (}a_{n,m}\left(z_{m}-z_{\infty }\right){\big )}+z_{\infty }\lim _{n\to \infty }\sum _{m=1}^{n}{\big (}a_{n,m}{\big )}=0+z_{\infty }\cdot 1=z_{\infty }}, which completes the proof.

Silverman–Toeplitz theorem

Formal statement

Proof

Proving 1.

Proving 2.

Citations

Further reading