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 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 ƒn : X → Y is a sequence of uniformly continuous functions converging uniformly to a function ƒ, then ƒ must be uniformly continuous. Both continuity and uniform continuity cases can be formulated in a more general context where Y {\displaystyle Y} (rather than being a metric space) is a uniform space, in which the notions of uniform convergence and uniform continuity can still be defined. However, in this setting, the use of nets or filters is preferred over sequences for greater generality.
Proof 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:
d Y ( f ( x ) , f ( y ) ) < ε , ∀ y ∈ U {\displaystyle d_{Y}(f(x),f(y))<\varepsilon ,\qquad \forall y\in U}
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:
d Y ( f N ( t ) , f ( t ) ) < ε 3 , ∀ t ∈ X {\displaystyle d_{Y}(f_{N}(t),f(t))<{\frac {\varepsilon }{3}},\qquad \forall t\in X}
Moreover, since fN is continuous on X by hypothesis, for every x there exists a neighbourhood U such that:
d Y ( f N ( x ) , f N ( y ) ) < ε 3 , ∀ y ∈ U {\displaystyle d_{Y}(f_{N}(x),f_{N}(y))<{\frac {\varepsilon }{3}},\qquad \forall y\in U}
In the final step, we apply the triangle inequality in the following way:
d Y ( f ( x ) , f ( y ) ) ≤ d Y ( f ( x ) , f N ( x ) ) + d Y ( f N ( x ) , f N ( y ) ) + d Y ( f N ( y ) , f ( y ) ) < ε 3 + ε 3 + ε 3 = ε , ∀ y ∈ U {\displaystyle {\begin{aligned}d_{Y}(f(x),f(y))&\leq d_{Y}(f(x),f_{N}(x))+d_{Y}(f_{N}(x),f_{N}(y))+d_{Y}(f_{N}(y),f(y))\\&<{\frac {\varepsilon }{3}}+{\frac {\varepsilon }{3}}+{\frac {\varepsilon }{3}}=\varepsilon ,\qquad \forall y\in U\end{aligned}}}
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