Exponential Sequence is Uniformly Convergent on Compact Sets

From ProofWiki
Jump to navigation Jump to search


Let $\EE = \sequence {E_n}$ denote the sequence of complex functions $E_n: \C \to \C$ defined as:

$\map {E_n} z = \paren {1 + \dfrac z n}^n$

Let $K$ be a compact subset of $\C$.

Then $\EE$ is uniformly convergent on $K$.


$\EE$ is Uniformly Bounded on an open space containing $K$

First, from Equivalence of Definitions of Complex Exponential Function we see that $\EE$ is pointwise convergent to $\exp$.

From Combination Theorem for Continuous Complex Functions, $E_n$ is continuous for each $n \in \N$.

From Compact Subspace of Metric Space is Bounded, $K$ is bounded by some real number $M \in \R$.


$M' = M + 1$
$U = \set {z \in \C: \cmod z < M'}$
$z \in U$
$n \in \N$

From the Triangle Inequality for Complex Numbers:

$\cmod {1 + \dfrac z n} \le 1 + \dfrac {\cmod z} n$

Thus, from Power Function is Strictly Increasing over Positive Reals: Natural Exponent:

$\cmod {1 + \dfrac z n}^n \le \paren {1 + \dfrac {\cmod z} n}^n$

Call this result $(1)$.


\(\ds \cmod z\) \(<\) \(\ds M'\)
\(\ds \leadsto \ \ \) \(\ds 1 + \frac {\cmod z} n\) \(<\) \(\ds 1 + \frac {M'} n\) Divide both sides by $n$, add $1$ to both sides
\(\text {(2)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \paren {1 + \frac {\cmod z} n}^n\) \(<\) \(\ds \paren {1 + \frac {M'} n}^n\) Power Function is Strictly Increasing over Positive Reals: Natural Exponent

Thus it is sufficient to show that $\paren {1 + \dfrac {M'} n}^n$ is bounded.

From Exponential Sequence is Eventually Increasing:

$\exists N \in \N: n \ge N \implies \paren {1 + \dfrac {M'} n}^n \le \paren {1 + \dfrac {M'} {n + 1} }^{n + 1}$

So, for $n \ge N$:

\(\ds \paren {1 + \frac {M'} n}^n\) \(\le\) \(\ds \lim_{n \mathop \to \infty} \paren {1 + \frac {M'} n}^n\) Limit of Bounded Convergent Sequence is Bounded
\(\ds \) \(=\) \(\ds \exp M'\)

Now, for each $n \in \set {1, 2, \ldots, N - 1}$:

$\exists M_n \in \R: \paren {1 + \dfrac {M'} n}^n \le M_n$

by Continuous Function on Compact Subspace of Euclidean Space is Bounded.


$\forall n \in \N: \paren {1 + \dfrac {M'} n}^n \le \max \set {M_1, M_2, \ldots, M_{N - 1}, \exp M'}$

Call this result $(3)$.

Hence, for $z \in U$; that is, for $\cmod z < {M'}$:

\(\ds \forall n \in \N: \, \) \(\ds \cmod {\paren {1 + \frac z n}^n}\) \(=\) \(\ds \cmod {1 + \dfrac z n}^n\) Complex Modulus of Product of Complex Numbers
\(\ds \) \(\le\) \(\ds \paren {1 + \dfrac {\cmod z} n}^n\) From $(1)$
\(\ds \) \(\le\) \(\ds \paren {1 + \frac {M'} n}^n\) From $(2)$
\(\ds \) \(\le\) \(\ds \max \set {M_1, M_2, \ldots, M_{N - 1}, \exp M'}\) From $(3)$

That is, $\EE$ is uniformly bounded on $U$.

$\EE$ is Uniformly Convergent on $K$

Note that, from Combination Theorem for Complex Derivatives:

$\forall n \in \N: \map {E_n} z$ is holomorphic on $U$.

From Uniformly Bounded Family is Locally Bounded, $\EE$ is locally bounded on $U$.

From Montel's Theorem, $\EE$ is a normal family.

From Vitali's Convergence Theorem, $\EE$ converges uniformly on any compact subset of $U$.

In particular, $\EE$ converges uniformly on $K$.

Hence the result.