Prove if F is Not Continuous Then Fn is Not Uniform Converge
In mathematics, the uniform limit theorem states that the uniform limit of any sequence of continuous functions is continuous.
Statement [edit]
More precisely, let X be a topological space, let Y be a metric space, and let ƒ n :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 n 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 n : [0, 1] →R be the sequence of functions ƒ n (x) =xn . Then each function ƒ n 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 adjacent image.
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 ƒ n :X →Y is a sequence of uniformly continuous functions converging uniformly to a function ƒ, then ƒ must be uniformly continuous.
Proof [edit]
In order to prove the continuity of f, we have to show that for every ε > 0, there exists a neighbourhood U of any point x of X such that:
Consider an arbitrary ε > 0. Since the sequence of functions (fn) converges uniformly to f by hypothesis, there exists a natural number N such that:
Moreover, since fN is continuous on X by hypothesis, for every x there exists a neighbourhood U such that:
In the final step, we apply the triangle inequality in the following way:
Hence, we have shown that the first inequality in the proof holds, so by definition f is continuous everywhere on X.
Uniform limit theorem in complex analysis [edit]
There are also variants of the uniform limit theorem that are used in complex analysis, albeit with modified assumptions.
Theorem.[1] Let be an open and connected subset of the complex numbers. Suppose that is a sequence of holomorphic functions that converges uniformly to a function on every compact subset of . Then is holomorphic in , and moreover, the sequence of derivatives converges uniformly to on every compact subset of .
Theorem.[2] Let be an open and connected subset of the complex numbers. Suppose that is a sequence of univalent[3] functions that converges uniformly to a function . Then is holomorphic, and moreover, is either univalent or constant in .
Notes [edit]
- ^ E.M.Stein, R.Shakarachi (2003), pp.53-54.
- ^ E.C.Titchmarsh (1939), p.200.
- ^ Univalent means holomorphic and injective.
References [edit]
- James Munkres (1999). Topology (2nd ed.). Prentice Hall. ISBN0-13-181629-2.
- E.M.Stein, R.Shakarachi (2003). Complex Analysis (Princeton Lectures in Analysis, No. 2), Princeton University Press, pp.53-54.
- E.C.Titchmarsh (1939). The Theory of Functions, 2002 Reprint, Oxford Science Publications.
Source: https://en.wikipedia.org/wiki/Uniform_limit_theorem