Uniformly bounded derivatives implies globally analytic: Difference between revisions

Latest revision as of 20:27, 12 July 2012

Statement

Global statement

Suppose $f$ is an infinitely differentiable function on $R$ such that, for any fixed $a < b \in R$ , there is a constant $C$ (possibly dependent on $a, b$ ) such that for all nonnegative integers $n$ , we have:

$| f^{(n)} (t) | \leq C \forall t \in [a, b]$

Then, $f$ is a globally analytic function: the Taylor series of $f$ about any point in $R$ converges to $f$ . In particular, the Taylor series of $f$ about 0 converges to $f$ .

Facts used

Max-estimate version of Lagrange formula

Examples

The functions $\exp, \sin, \cos$ all fit this description.

If $f = \exp$ , we know that each of the derivatives equals $\exp$ , so $f^{(n)} (t) = f (t)$ for all $t \in [a, b]$ . Since $\exp$ is continuous, it is bounded on the closed interval $[a, b]$ , and the upper bound for $\exp$ thus serves as a uniform bound for all its derivatives. (In fact, since $f$ is increasing, we can explicitly take $C = \exp (b)$ ).

For $f = \sin$ or $f = \cos$ , we know that all the derivatives are $\pm \sin$ or $\pm \cos$ , so their magnitude is at most 1. Thus, we can take $C = 1$ .

Proof

Given: $f$ is an infinitely differentiable function on $R$ such that, for any fixed $a, b \in R$ , there is a constant $C$ (possibly dependent on $a, b$ ) such that for all nonnegative integers $n$ , we have:

$| f^{(n)} (t) | \leq C \forall t \in [a, b]$

A point $x_{0} \in R$ and a point $x \in R$ .

To prove: The Taylor series of $f$ at $x_{0}$ , evaluated at $x$ , converges to $f (x)$ .

Proof: Note that if $x_{0} = x$ , there is nothing to prove, so we consider the case $x \neq x_{0}$ .

In order to show this, it suffices to show that $lim_{n \to \infty} P_{n} (f; x_{0}) (x) = f (x)$ where $P_{n} (f; x_{0}) (x)$ denotes the $n^{t h}$ Taylor polynomial of $f$ at $x_{0}$ , evaluated at $x$ .

This in turn is equivalent to showing that the remainder approaches zero:

'Want to show: $lim_{n \to \infty} R_{n} (f; x_{0}) (x) = 0$

where $R_{n} (f; x_{0}) (x) = f (x) - P_{n} (f; x_{0}) (x)$ .

Proof of what we want to show: By Fact (1), we have that:

$| R_{n} (f; x_{0}) (x) | \leq ({max}_{t \in J} | f^{(n + 1)} (t) |) \frac{| x - x_{0} |^{n + 1}}{(n + 1)!}$

where $J$ is the interval joining $x_{0}$ to $x$ . Let $a = min {x, x_{0}}$ and $b = max {x, x_{0}}$ . The interval $J$ is the interval $[a, b]$ .

Now, from the given data, there exists $C$ , dependent on $x$ and $x_{0}$ but not on $n$ , such that:

${max}_{t \in J} | f^{(n + 1)} (t) | \leq C \forall n$

Plugging this in, we get that:

$| R_{n} (f; x_{0}) (x) | \leq C \frac{| x - x_{0} |^{n + 1}}{(n + 1)!}$

Now taking the limit as $n \to \infty$ , we get:

$lim_{n \to \infty} | R_{n} (f; x_{0}) (x) | \leq C lim_{n \to \infty} \frac{| x - x_{0} |^{n + 1}}{(n + 1)!}$

Since exponentials grow faster than power functions, the expression under the limit goes to zero, so we are left with a right side of zero, hence the left side limit is zero, and we are done.

@@ Line 3: / Line 3: @@
 ===Global statement===
-Suppose <math>f</math> is an infinitely differentiable function on <math>\R</math> such that, for any fixed <math>a,b \in \R</math>, there is a constant <math>C</math> (possibly dependent on <math>a,b</math>) such that for all nonnegative integers <math>n</math>, we have:
+Suppose <math>f</math> is an infinitely differentiable function on <math>\R</math> such that, for any fixed <math>a<b \in \R</math>, there is a constant <math>C</math> (possibly dependent on <math>a,b</math>) such that for all nonnegative integers <math>n</math>, we have:
-<math>|f^{(n)}(x)| \le C \ \forall x \in [a,b]</math>
+<math>|f^{(n)}(t)| \le C \ \forall t \in [a,b]</math>
 Then, <math>f</math> is a [[globally analytic function]]: the [[Taylor series]] of <math>f</math> about any point in <math>\R</math> converges to <math>f</math>. In particular, the Taylor series of <math>f</math> about 0 converges to <math>f</math>.
+==Facts used==
+# [[uses::Max-estimate version of Lagrange formula]]
 ==Examples==
 The functions <math>\exp, \sin, \cos</math> all fit this description.
+If <math>f = \exp</math>, we know that each of the derivatives equals <math>\exp</math>, so <math>f^{(n)}(t) = f(t)</math> for all <math>t \in [a,b]</math>. Since <math>\exp</math> is continuous, it is bounded on the closed interval <math>[a,b]</math>, and the upper bound for <math>\exp</math> thus serves as a uniform bound for all its derivatives. (In fact, since <math>f</math> is increasing, we can explicitly take <math>C = \exp(b)</math>).
+For <math>f = \sin</math> or <math>f = \cos</math>, we know that all the derivatives are <math>\pm \sin</math> or <math>\pm \cos</math>, so their magnitude is at most 1. Thus, we can take <math>C = 1</math>.
+==Proof==
+'''Given''': <math>f</math> is an infinitely differentiable function on <math>\R</math> such that, for any fixed <math>a,b \in \R</math>, there is a constant <math>C</math> (possibly dependent on <math>a,b</math>) such that for all nonnegative integers <math>n</math>, we have:
+<math>|f^{(n)}(t)| \le C \ \forall t \in [a,b]</math>
+A point <math>x_0 \in \R</math> and a point <math>x \in \R</math>.
+'''To prove''': The Taylor series of <math>f</math> at <math>x_0</math>, evaluated at <math>x</math>, converges to <math>f(x)</math>.
+'''Proof''': Note that if <math>x_0 = x</math>, there is nothing to prove, so we consider the case <math>x \ne x_0</math>.
+In order to show this, it suffices to show that <math>\lim_{n \to \infty} P_n(f;x_0)(x) = f(x)</math> where <math>P_n(f;x_0)(x)</math> denotes the <math>n^{th}</math> Taylor polynomial of <math>f</math> at <math>x_0</math>, evaluated at <math>x</math>.
+This in turn is equivalent to showing that the ''remainder'' approaches zero:
+''''Want to show''': <math>\lim_{n \to \infty} R_n(f;x_0)(x) = 0</math>
+where <math>R_n(f;x_0)(x) = f(x) - P_n(f;x_0)(x)</math>.
+'''Proof of what we want to show''': By Fact (1), we have that:
+<math>|R_n(f;x_0)(x)| \le \left(\max_{t \in J} |f^{(n+1)}(t)|\right) \frac{|x - x_0|^{n+1}}{(n + 1)!}</math>
+where <math>J</math> is the interval joining <math>x_0</math> to <math>x</math>. Let <math>a = \min \{ x,x_0 \}</math> and <math>b = \max \{ x, x_0 \}</math>. The interval <math>J</math> is the interval <math>[a,b]</math>.
+Now, from the given data, there exists <math>C</math>, dependent on <math>x</math> and <math>x_0</math> but ''not'' on <math>n</math>, such that:
+<math>\max_{t \in J} |f^{(n+1)}(t)| \le C \ \forall \ n</math>
+Plugging this in, we get that:
+<math>|R_n(f;x_0)(x)| \le C \frac{|x - x_0|^{n+1}}{(n + 1)!}</math>
+Now taking the limit as <math>n \to \infty</math>, we get:
+<math>\lim_{n \to \infty} |R_n(f;x_0)(x)| \le C \lim_{n \to \infty} \frac{|x - x_0|^{n+1}}{(n + 1)!}</math>
+Since exponentials grow faster than power functions, the expression under the limit goes to zero, so we are left with a right side of zero, hence the left side limit is zero, and we are done.