Theory of solar cells
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
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