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Cesàro Mean

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Theorem

Let $\sequence {a_n}$ be a sequence of complex numbers.

Suppose that $\sequence {a_n}$ converges to $\ell$ in $\C$:

$\ds \lim_{n \mathop \to \infty} a_n = \ell$


Then also:

$\ds \lim_{n \mathop \to \infty} \frac {a_1 + \dotsb + a_n} n = \ell$


Proof

For every fixed integer $n_0$, we write:

$\ds \cmod {\frac {a_1 + \dotsb + a_n} n - \ell} \le \frac {\cmod {a_1 - \ell} + \dotsb + \cmod {a_n - \ell} } n \le \frac {n_0 \ds \sup_{k \mathop \le n_0} \cmod {a_k - \ell} } n + \sup_{n_0 \mathop < k \mathop \le n} \cmod {a_k - \ell}$

As $n$ tends to $+\infty$, we get:

$\ds \limsup_{n \mathop \to \infty} \cmod {\frac {a_1 + \dotsb + a_n} n - \ell} \le \sup_{k \mathop > n_0} \cmod {a_k - \ell}$

As $n_0$ tends to $+\infty$, we finally conclude:

$\ds \limsup_{n \mathop \to \infty} \cmod {\frac {a_1 + \dotsb + a_n} n - \ell} = 0$

$\blacksquare$


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Remarks


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In particular: Need to split off the case for normed vector spaces
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  • When working with an arbitrary sequence $\sequence {a_n}$ of real numbers, the same truncation trick leads to:
$\ds \liminf_{n \mathop \to \infty} a_n \le \liminf_{n \mathop \to \infty} \frac {a_1 + \dotsb + a_n} n \le \limsup_{n \mathop \to \infty} \frac {a_1 + \dotsb + a_n} n \le \limsup_{n \mathop \to \infty} a_n$

As a corollary, the conclusion of the theorem holds in the real case when $\ell = \pm \infty$.


Source of Name

This entry was named for Ernesto Cesàro.

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