Theory of solar cells: Difference between revisions
en>FrescoBot m Bot: standard sections headers |
|||
Line 1: | Line 1: | ||
[[Image:Drini nonuniformconvergence SVG.svg|thumb|350px|right|Counterexample to a strengthening of the uniform limit theorem, in which pointwise convergence, rather than uniform convergence, is assumed. The continuous green functions <math>\scriptstyle \scriptstyle\sin^n(x)</math> converge to the non-continuous red function. This can happen only if convergence is not uniform.]] | |||
In [[mathematics]], the '''uniform limit theorem''' states that the [[uniform convergence|uniform limit]] of any sequence of [[continuous function]]s is continuous. | |||
==Statement== | |||
More precisely, let ''X'' be a [[topological space]], let ''Y'' be a [[metric space]], and let ƒ<sub>''n''</sub> : ''X'' → ''Y'' be a sequence of functions converging uniformly to a function ƒ : ''X'' → ''Y''. According to the uniform limit theorem, if each of the functions ƒ<sub>''n''</sub> is continuous, then the limit ƒ must be continuous as well. | |||
This theorem does not hold if uniform convergence is replaced by [[pointwise convergence]]. For example, let ƒ<sub>''n''</sub> : [0, 1] → '''R''' be the sequence of functions ƒ<sub>''n''</sub>(''x'') = ''x<sup>n</sub>''. Then each function ƒ<sub>''n''</sub> is continuous, but the sequence converges pointwise to the discontinuous function ƒ that is zero on [0, 1) but has ƒ(1) = 1. Another example is shown in the image to the right. | |||
In terms of [[function space]]s, the uniform limit theorem says that the space ''C''(''X'', ''Y'') of all continuous functions from a topological space ''X'' to a metric space ''Y'' is a [[closed set|closed subset]] of ''Y<sup>X</sup>'' under the [[uniform metric]]. In the case where ''Y'' is [[Complete metric space|complete]], it follows that ''C''(''X'', ''Y'') is itself a complete metric space. In particular, if ''Y'' is a [[Banach space]], then ''C''(''X'', ''Y'') is itself a Banach space under the [[uniform norm]]. | |||
The uniform limit theorem also holds if continuity is replaced by [[uniform continuity]]. That is, if ''X'' and ''Y'' are metric spaces and ƒ<sub>''n''</sub> : ''X'' → ''Y'' is a sequence of uniformly continuous functions converging uniformly to a function ƒ, then ƒ must be uniformly continuous. | |||
==Proof== | |||
The uniform limit theorem is proved by the "''ε''/3 trick", and is the archetypal example of this trick. | |||
More precisely we want to prove that, for every ''ε'' > 0, there is a [[neighbourhood (mathematics)|neighbourhood]] ''N'' of any point ''x'' of ''X'' such that ''d<sub>Y</sub>''(''f''<sub>0</sub>(''x''),''f''<sub>0</sub>(''y'')) < ''ε'' whenever ''y'' is in ''N''. In order to do so we make use of the [[triangle inequality]] applied to the metric ''d<sub>Y</sub>'' on ''Y'' in the following way | |||
:<math>d_Y(f_0(x),f_0(y))\leq d_Y(f_0(x),f_n(x))+d_Y(f_n(x),f_n(y))+d_Y(f_n(y),f_0(y)).</math> | |||
Let us fix an arbitrary ''ε'' > 0. Since the sequence of functions {''f<sub>n</sub>''} converges uniformly to ''f'' by hypothesis then we may always choose a natural number ''n''<sub>0</sub> such that ''n'' > ''n''<sub>0</sub> implies | |||
:<math>d_Y(f_n(t),f_0(t))<\frac\epsilon3,\qquad\forall t\in X.</math> | |||
Moreover, since each ''f<sub>n</sub>'' is continuous on the whole ''X'' by hypothesis, there is a neighbourhood ''N'' of any ''x'' such that ''d<sub>Y</sub>''(''f''<sub>n</sub>(''x''),''f''<sub>n</sub>(''y'')) < ''ε''/3 whenever ''y'' is in ''N''. Taking the limit on ''n'' to ∞ on the above triangular inequality we then obtain that, for every ''ε'' > 0, there is a neighbourhood ''N'' of any point ''x'' of ''X'' such that | |||
:<math>d_Y(f_0(x),f_0(y))\leq \frac\epsilon3+\frac\epsilon3+\frac\epsilon3=\epsilon</math> | |||
whenever ''y'' is in ''N''. Hence ''f'' is continuous everywhere on ''X'' by definition. | |||
==References== | |||
* {{cite book | |||
| author = [[James Munkres]] | |||
| year = 1999 | |||
| title = Topology | |||
| edition = 2nd edition | |||
| publisher = [[Prentice Hall]] | |||
| isbn = 0-13-181629-2 | |||
}} | |||
[[Category:Theorems in real analysis]] | |||
[[Category:Topology of function spaces]] |
Revision as of 03:05, 12 January 2014
In mathematics, the uniform limit theorem states that the uniform limit of any sequence of continuous functions is continuous.
Statement
More precisely, let X be a topological space, let Y be a metric space, and let ƒn : X → Y be a sequence of functions converging uniformly to a function ƒ : X → Y. According to the uniform limit theorem, if each of the functions ƒn is continuous, then the limit ƒ must be continuous as well.
This theorem does not hold if uniform convergence is replaced by pointwise convergence. For example, let ƒn : [0, 1] → R be the sequence of functions ƒn(x) = xn. Then each function ƒn is continuous, but the sequence converges pointwise to the discontinuous function ƒ that is zero on [0, 1) but has ƒ(1) = 1. Another example is shown in the image to the right.
In terms of function spaces, the uniform limit theorem says that the space C(X, Y) of all continuous functions from a topological space X to a metric space Y is a closed subset of YX under the uniform metric. In the case where Y is complete, it follows that C(X, Y) is itself a complete metric space. In particular, if Y is a Banach space, then C(X, Y) is itself a Banach space under the uniform norm.
The uniform limit theorem also holds if continuity is replaced by uniform continuity. That is, if X and Y are metric spaces and ƒn : X → Y is a sequence of uniformly continuous functions converging uniformly to a function ƒ, then ƒ must be uniformly continuous.
Proof
The uniform limit theorem is proved by the "ε/3 trick", and is the archetypal example of this trick.
More precisely we want to prove that, for every ε > 0, there is a neighbourhood N of any point x of X such that dY(f0(x),f0(y)) < ε whenever y is in N. In order to do so we make use of the triangle inequality applied to the metric dY on Y in the following way
Let us fix an arbitrary ε > 0. Since the sequence of functions {fn} converges uniformly to f by hypothesis then we may always choose a natural number n0 such that n > n0 implies
Moreover, since each fn is continuous on the whole X by hypothesis, there is a neighbourhood N of any x such that dY(fn(x),fn(y)) < ε/3 whenever y is in N. Taking the limit on n to ∞ on the above triangular inequality we then obtain that, for every ε > 0, there is a neighbourhood N of any point x of X such that
whenever y is in N. Hence f is continuous everywhere on X by definition.
References
- 20 year-old Real Estate Agent Rusty from Saint-Paul, has hobbies and interests which includes monopoly, property developers in singapore and poker. Will soon undertake a contiki trip that may include going to the Lower Valley of the Omo.
My blog: http://www.primaboinca.com/view_profile.php?userid=5889534