Tauber’s theorem and Karamata’s proof of the Hardy-Littlewood tauberian theorem

The following lemma is attributed to Kronecker by Knopp.

Lemma 1 (Kronecker’s lemma).

If $b_n\to 0$ then

$$\frac{b_0+b_1+\mathrm{\cdots }+b_n}{n+1}\to 0.$$

Proof.

Suppose that $|b_n|\le K$ for all $n$, and let $ϵ>0$. As $b_n\to 0$ there is some $n_0$ such that $n\ge n_0$ implies that . If $n\ge \frac{(n_0+1)K}{ϵ}$, then

$\left|{\displaystyle \frac{b_0+b_1+\mathrm{\cdots }+b_n}{n+1}}\right|$ $\le \left|{\displaystyle \frac{b_0+b_1+\mathrm{\cdots }+b_{n_0}}{n+1}}\right|+\left|{\displaystyle \frac{b_{n_0}+\mathrm{\cdots }+b_n}{n+1}}\right|$

$\le {\displaystyle \frac{(n_0+1)K}{n+1}}+{\displaystyle \frac{(n-n_0)ϵ}{n+1}}$

$\le ϵ+ϵ.$

∎

We now use the above lemma to prove Tauber’s theorem.

Theorem 2 (Tauber’s theorem).

If $a_n=o(1/n)$ and ${\sum }_{n=0}^{\mathrm{\infty }}{a}_n{x}^n\to s$ as $x\to 1^{-}$, then

$$\sum _{n=0}^{\mathrm{\infty }}{a}_n=s.$$

Proof.

Let $ϵ>0$. Because ${\sum }_{n=0}^{\mathrm{\infty }}{a}_n{x}^n\to s$ as $x\to 1^{-}$, there is some $\delta >0$ such that $x>1-\delta$ implies that

Next, because $n|a_n|\to 0$, there is some $N>\frac{1}{\delta }$ such that (i) if $n\ge N$ then and by Lemma 1, (ii) .

Take $x=1-\frac{1}{N}$, so $N=\frac{1}{1-x}$ and $1-x=\frac{1}{N}$. We have

$\left|{\displaystyle \sum _{n=N+1}^{\mathrm{\infty }}}{a}_n{x}^n\right|$ $=\left|{\displaystyle \sum _{n=N+1}^{\mathrm{\infty }}}n{a}_n\cdot {\displaystyle \frac{x^n}{n}}\right|$

$=ϵ\cdot {\displaystyle \frac{N}{N+1}}$

Also, using

we have

$\left|{\displaystyle \sum _{n=0}^N}{a}_n(1-x^n)\right|$ $\le {\displaystyle \sum _{n=0}^N}|a_n|(1-x^n)$

$={\displaystyle \sum _{n=0}^N}{\displaystyle \frac{|a_n|n}{N}}$

$={\displaystyle \frac{N+1}{N}}\cdot {\displaystyle \frac{1}{N+1}}{\displaystyle \sum _{n=0}^N}n|a_n|$

Now,

${\displaystyle \sum _{n=0}^N}{a}_n-s$ $={\displaystyle \sum _{n=0}^N}{a}_n-{\displaystyle \sum _{n=0}^N}{a}_n{x}^n+{\displaystyle \sum _{n=0}^N}{a}_n{x}^n-s$

$={\displaystyle \sum _{n=0}^N}{a}_n(1-x^n)+{\displaystyle \sum _{n=0}^N}{a}_n{x}^n-s$

$={\displaystyle \sum _{n=0}^N}{a}_n(1-x^n)+{\displaystyle \sum _{n=0}^N}{a}_n{x}^n+{\displaystyle \sum _{n=N+1}^{\mathrm{\infty }}}{a}_n{x}^n-{\displaystyle \sum _{n=N+1}^{\mathrm{\infty }}}{a}_n{x}^n-s$

$={\displaystyle \sum _{n=0}^N}{a}_n(1-x^n)+{\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{a}_n{x}^n-s-{\displaystyle \sum _{n=N+1}^{\mathrm{\infty }}}{a}_n{x}^n$

and then

$\left|{\displaystyle \sum _{n=0}^N}{a}_n-s\right|$ $\le \left|{\displaystyle \sum _{n=0}^N}{a}_n(1-x^n)\right|+\left|{\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{a}_n{x}^n-s\right|+\left|{\displaystyle \sum _{n=N+1}^{\mathrm{\infty }}}{a}_n{x}^n\right|$

proving the claim. ∎

Lemma 3.

Let $g:[0,1]\to ℝ$ and . Suppose that the restrictions of $g$ to $[0,c)$ and $[c,1]$ are continuous and that

$$g(c-0)=\underset{x\to c^{-}}{lim}g(x)\le g(c).$$

For $ϵ>0$, there are are polynomials $p(x)$ and $P(x)$ such that

$$p(x)\le g(x)\le P(x),0\le x\le 1$$

and

$${\parallel g-p\parallel }_1\le ϵ,{\parallel g-P\parallel }_1\le ϵ.$$

Proof.

There is some $\delta >0$ such that implies that

$$g(c-0)-\frac{ϵ}{2}\le g(x)\le g(c-0)+\frac{ϵ}{2};$$

further, take and .

Take $L$ to be the linear function satisfying

$$L(c-\delta )=g(c-\delta )+\frac{ϵ}{2},L(c)=g(c)+\frac{ϵ}{2}.$$

For ,

$L(x)-g(x)$ $=L(x)-g(c-\delta )+g(c-\delta )-g(c-0)+g(c-0)-g(x)$

$=L(x)-L(c-\delta )+{\displaystyle \frac{ϵ}{2}}+g(c-\delta )-g(c-0)+g(c-0)-g(x)$

$\le L(c)-L(c-\delta )+{\displaystyle \frac{ϵ}{2}}+{\displaystyle \frac{ϵ}{2}}+{\displaystyle \frac{ϵ}{2}}$

$=g(c)-g(c-\delta )+{\displaystyle \frac{3ϵ}{2}}$

$=g(c)-g(c-0)+g(c-0)-g(c-\delta )+{\displaystyle \frac{3ϵ}{2}}$

Define $\mathrm{\Phi }:[0,1]\to ℝ$ by

$\mathrm{\Phi }$ is continuous and $\mathrm{\Phi }\ge g+\frac{ϵ}{2}$. We have

${\parallel g-\mathrm{\Phi }\parallel }_1$ $={\displaystyle {\int }_0^1}(\mathrm{\Phi }(x)-g(x))𝑑x$

$={\displaystyle {\int }_0^{c-\delta }}{\displaystyle \frac{ϵ}{2}}𝑑x+{\displaystyle {\int }_{c-\delta }^c}(\mathrm{\Phi }(x)-g(x))𝑑x+{\displaystyle {\int }_c^1}{\displaystyle \frac{ϵ}{2}}𝑑x$

$\le {\displaystyle \frac{ϵ}{2}}+{\displaystyle {\int }_{c-\delta }^c}\max \{L(x)-g(x),{\displaystyle \frac{ϵ}{2}}\}𝑑x$

$\le {\displaystyle \frac{ϵ}{2}}+{\displaystyle {\int }_{c-\delta }^c}\max \{{\displaystyle \frac{2ϵ}{\delta }},{\displaystyle \frac{ϵ}{2}}\}𝑑x$

$={\displaystyle \frac{ϵ}{2}}+\delta \cdot {\displaystyle \frac{2ϵ}{\delta }}$

$={\displaystyle \frac{5ϵ}{2}}.$

Because $\mathrm{\Phi }$ is continuous, by the Weierstrass approximation theorem there is a polynomial $P(x)$ such that ${\parallel \mathrm{\Phi }-P\parallel }_{\mathrm{\infty }}\le \frac{ϵ}{2}$. Then,

$$g(x)\le P(x),0\le x\le 1,$$

and

On the other hand, take $l$ to be the linear function satisfying

$$l(c-\delta )=g(c-\delta )-\frac{ϵ}{2},l(c)=g(c)-\frac{ϵ}{2}.$$

One checks that for .

Define $\varphi :[0,1]\to ℝ$ by

which is continuous and satisfies $\varphi \le g-\frac{ϵ}{2}$. One checks that

Because $\varphi$ is continuous, there is a polynomial $p(x)$ such that ${\parallel \varphi -p\parallel }_{\mathrm{\infty }}\le \frac{ϵ}{2}$. Then,

$$p(x)\le g(x),0\le x\le 1,$$

and

∎

The following is the Hardy-Littlewood tauberian theorem.

Theorem 4 (Hardy-Littlewood tauberian theorem).

If $a_n\ge 0$ for all $n$ and

$$\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n\sim \frac{1}{1-x},x\to 1^{-},$$

then

$$s_n=\sum _{\nu =0}^n{a}_{\nu }\sim n.$$

Proof.

For any $k\ge 0$,

$(1-x){\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{a}_n{x}^n{(x^n)}^k$ $={\displaystyle \frac{1-x}{1-x^{k+1}}}(1-x^{k+1}){\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{a}_n{(x^{k+1})}^n$

$={\displaystyle \frac{1}{1+x+\mathrm{\cdots }+x^k}}(1-x^{k+1}){\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{a}_n{(x^{k+1})}^n$

$\to {\displaystyle \frac{1}{k+1}}\cdot 1$

$={\displaystyle {\int }_0^1}{t}^k𝑑t,$

as $x\to 1^{-}$. Hence if $P(x)$ is a polynomial, then

$$\underset{x\to 1^{-}}{lim}(1-x)\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^{n}P(x^n)={\int }_0^{1}P(t)𝑑t.$$ (1)

Define $g:[0,1]\to ℝ$ by

Let $ϵ>0$. By Lemma 3, there are polynomials $p(x),P(x)$ such that

$$p(x)\le g(x)\le P(x),0\le x\le 1$$

and

$${\parallel g-p\parallel }_1\le ϵ,{\parallel P-g\parallel }_1\le ϵ.$$

Because the coefficients $a_n$ are nonnegative, taking upper limits and then using (1) we obtain

$\underset{x\to 1^{-}}{lim\; sup}(1-x){\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{a}_n{x}^{n}g(x^n)$ $\le \underset{x\to 1^{-}}{lim\; sup}(1-x){\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{a}_n{x}^{n}P(x^n)$

$=\underset{x\to 1^{-}}{lim}(1-x){\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{a}_n{x}^{n}P(x^n)$

$={\displaystyle {\int }_0^1}P(t)𝑑t$

Taking lower limits and then using (1) we obtain

$\underset{x\to 1^{-}}{lim\; inf}(1-x){\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{a}_n{x}^{n}g(x^n)$ $\ge \underset{x\to 1^{-}}{lim\; inf}(1-x){\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{a}_n{x}^{n}p(x^n)$

$=\underset{x\to 1^{-}}{lim}(1-x){\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{a}_n{x}^{n}p(x^n)$

$={\displaystyle {\int }_0^1}p(t)𝑑t$

$>{\displaystyle {\int }_0^1}g(t)𝑑t-ϵ.$

The above two inequalities do not depend on the polynomials $p(x),P(x)$ but only on $ϵ$, and taking $ϵ\to 0$ yields

$$\underset{x\to 1^{-}}{lim\; sup}(1-x)\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^{n}g(x^n)\le {\int }_0^{1}g(t)𝑑t$$

and

$$\underset{x\to 1^{-}}{lim\; inf}(1-x)\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^{n}g(x^n)\ge {\int }_0^{1}g(t)𝑑t.$$

Thus

$$\underset{x\to 1^{-}}{lim}(1-x)\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^{n}g(x^n)={\int }_0^{1}g(t)𝑑t={\int }_{e^{-1}}^1{t}^{-1}𝑑t=1.$$ (2)

For $x=e^{-1/N}$ we have

${\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{a}_n{x}^{n}g(x^n)$ $={\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{a}_n{e}^{-n/N}g(e^{-n/N})$

$={\displaystyle \sum _{n=0}^N}{a}_n{e}^{-n/N}{e}^{n/N}$

$=s_N.$

Thus, (2) tells us that

$$\underset{N\to \mathrm{\infty }}{lim}(1-e^{-1/N})s_N=1.$$

That is,

$$s_N\sim \frac{1}{1-e^{-1/N}},$$

and using

$$\frac{1}{1-e^{-1/N}}=N+\frac{1}{2}+O(N^{-1})$$

we get

$$s_N\sim N,$$

completing the proof. ∎