Difference between revisions of "Uniformly bounded derivatives implies globally analytic"

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(Global statement)
 
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==Statement==
 
==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:
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===Global statement===
  
<math>|f^{(n)}(x)| \le C \ \forall x \in [a,b]</math>
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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:
  
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 0.
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<math>|f^{(n)}(t)| \le C \ \forall t \in [a,b]</math>
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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>.
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==Facts used==
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# [[uses::Max-estimate version of Lagrange formula]]
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==Examples==
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The functions <math>\exp, \sin, \cos</math> all fit this description.
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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>).
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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>.
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==Proof==
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'''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:
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<math>|f^{(n)}(t)| \le C \ \forall t \in [a,b]</math>
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A point <math>x_0 \in \R</math> and a point <math>x \in \R</math>.
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'''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>.
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'''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>.
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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>.
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This in turn is equivalent to showing that the ''remainder'' approaches zero:
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''''Want to show''': <math>\lim_{n \to \infty} R_n(f;x_0)(x) = 0</math>
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where <math>R_n(f;x_0)(x) = f(x) - P_n(f;x_0)(x)</math>.
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'''Proof of what we want to show''': By Fact (1), we have that:
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<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>
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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>.
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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:
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<math>\max_{t \in J} |f^{(n+1)}(t)| \le C \ \forall \ n</math>
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Plugging this in, we get that:
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<math>|R_n(f;x_0)(x)| \le C \frac{|x - x_0|^{n+1}}{(n + 1)!}</math>
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Now taking the limit as <math>n \to \infty</math>, we get:
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<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>
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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.

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)| \le 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

  1. 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)| \le 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 \ne 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^{th} 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)| \le \left(\max_{t \in J} |f^{(n+1)}(t)|\right) \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)| \le C \ \forall \ n

Plugging this in, we get that:

|R_n(f;x_0)(x)| \le 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)| \le 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.