Uniformly Convergent Sequence Evaluated on Convergent Sequence

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Theorem

Let $X = \struct {A, d_X}$ and $Y = \struct {B, d_Y}$ be metric spaces.

Let $K$ be a subspace of $X$.

Let $f: X \to Y$ be a mapping.

Let $\FF = \sequence {f_n}$ be a sequence of continuous mappings $f_n: X \to Y$ that converges to $f$ uniformly on $K$.

Let $\sequence {a_n}$ be a convergent sequence in $K$ with limit $a \in K$.


Then the sequence $\sequence {\map {f_n} {a_n} }$ is convergent such that:

$\ds \lim_{n \mathop \to \infty} \map {f_n} {a_n} = \map f a$


Proof

We want to show that:

$\map {d_Y} {\map {f_n} {a_n} , \map f a} \to 0$ as $n \to \infty$


Let $\epsilon \in \R_{>0}$ be fixed.

From the Triangle Equality, it follows that:

$\map {d_Y} {\map {f_n} {a_n} , \map f a} \le \map {d_Y} {\map {f_n} {a_n} , \map f {a_n} } + \map {d_Y} {\map f {a_n} , \map f a}$


From the Uniform Limit Theorem, it follows that $f$ is continuous.

Since $a_n \to a$, Sequential Continuity is Equivalent to Continuity in Metric Space tells us that:

$\exists M \in \N : \forall n \ge M : \map {d_Y} {\map f {a_n} , \map f a} < \dfrac \epsilon 2$


Since $\sequence {f_n}$ is uniformly convergent on $K$, it follows that:

$\exists N \in \N : \forall n \ge N : \map {d_Y} {\map {f_n} {a_n} , \map f {a_n} } < \dfrac \epsilon 2$


So if $n \ge \max \set {M, N}$, it follows that:

\(\ds \map {d_Y} {\map {f_n} {a_n} , \map f a}\) \(\le\) \(\ds \map {d_Y} {\map {f_n} {a_n} , \map f {a_n} } + \map {d_Y} {\map f {a_n} , \map f a}\)
\(\ds \) \(<\) \(\ds \dfrac \epsilon 2 + \dfrac \epsilon 2\)
\(\ds \) \(=\) \(\ds \epsilon\)

which shows that $\map {d_Y} {\map {f_n} {a_n} , \map f a} \to 0$ as $n \to \infty$.

$\blacksquare$