- #1
mattmns
- 1,121
- 5
Here is the question from our book:
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Let (X,d) be a complete metric space, and let f:X\to X and g:X\to X be two strict contractions on X with contraction coefficients c and c' respectively. From the Contraction Mapping Theorem we know that f has some fixed point x_0, and g has some fixed point y_0. Suppose we know that there is an \epsilon > 0 such that d(f(x),g(x)) \leq \epsilon for all x \in X. Show that d(x_0,y_0) \leq \epsilon/(1-\max(c,c')). Thus nearby contractions have nearby fixed points.
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Below is my proof. I believe everything I have is correct. My question is why do we not conclude that d(x_0,y_0) < \epsilon/(1-\min(c,c')) as I believe this is a stronger result that follows (with very slight modification) from my proof below.
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Proof.
Let m = \max(c,c').
So we have d(f(x),f(y)) < md(x,y) and d(g(x),g(y)) < md(x,y) for all x,y \in X.
Since x_0 is a fixed point of f, and y_0 is a fixed point of g, we know that f(x_0) = x_0, and g(y_0) = y_0. Hence we get the following
<br /> \begin{align*}<br /> d(x_0,y_0) & = d(f(x_0),g(y_0)) \\<br /> & \leq d(f(x_0),f(y_0)) + d(f(y_0),g(y_0)) \\<br /> & \leq md(x_0,y_0) + \epsilon<br /> \end{align*}<br />
So d(x_0,y_0) \leq md(x_0,y_0) + \epsilon.
Hence, d(x_0,y_0)(1 - m) \leq \epsilon.
Therefore d(x_0,y_0) \leq \dfrac{\epsilon}{1 - \max(c,c')}.
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I can't see anything wrong, but there is probably some subtlety that I am missing. Any ideas? Thanks!Definitions if needed.
Strict contraction: Let f:X\to X be a map, then f is a strict contraction if d(f(x),f(y) < d(x,y).
Fixed point: x is a fixed point of a function f if f(x) = x.
Contraction mapping theorem. Let (X,d) be a metric space, and let f:X\to X be a strict contraction. Then f can have at most one fixed point. Moreover, if we assume X is complete and non-empty then f has exactly one fixed point.
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Let (X,d) be a complete metric space, and let f:X\to X and g:X\to X be two strict contractions on X with contraction coefficients c and c' respectively. From the Contraction Mapping Theorem we know that f has some fixed point x_0, and g has some fixed point y_0. Suppose we know that there is an \epsilon > 0 such that d(f(x),g(x)) \leq \epsilon for all x \in X. Show that d(x_0,y_0) \leq \epsilon/(1-\max(c,c')). Thus nearby contractions have nearby fixed points.
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Below is my proof. I believe everything I have is correct. My question is why do we not conclude that d(x_0,y_0) < \epsilon/(1-\min(c,c')) as I believe this is a stronger result that follows (with very slight modification) from my proof below.
-------
Proof.
Let m = \max(c,c').
So we have d(f(x),f(y)) < md(x,y) and d(g(x),g(y)) < md(x,y) for all x,y \in X.
Since x_0 is a fixed point of f, and y_0 is a fixed point of g, we know that f(x_0) = x_0, and g(y_0) = y_0. Hence we get the following
<br /> \begin{align*}<br /> d(x_0,y_0) & = d(f(x_0),g(y_0)) \\<br /> & \leq d(f(x_0),f(y_0)) + d(f(y_0),g(y_0)) \\<br /> & \leq md(x_0,y_0) + \epsilon<br /> \end{align*}<br />
So d(x_0,y_0) \leq md(x_0,y_0) + \epsilon.
Hence, d(x_0,y_0)(1 - m) \leq \epsilon.
Therefore d(x_0,y_0) \leq \dfrac{\epsilon}{1 - \max(c,c')}.
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I can't see anything wrong, but there is probably some subtlety that I am missing. Any ideas? Thanks!Definitions if needed.
Strict contraction: Let f:X\to X be a map, then f is a strict contraction if d(f(x),f(y) < d(x,y).
Fixed point: x is a fixed point of a function f if f(x) = x.
Contraction mapping theorem. Let (X,d) be a metric space, and let f:X\to X be a strict contraction. Then f can have at most one fixed point. Moreover, if we assume X is complete and non-empty then f has exactly one fixed point.
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