Uniform Limit Theorem

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Theorem

Let $\left({M, d_M}\right)$ and $\left({N, d_N}\right)$ be metric spaces.

Let $\left \langle{f_n}\right \rangle$ be a sequence of mappings from $M$ to $N$ such that:

$(1): \quad \forall n \in \N: f_n$ is continuous at every point of $M$
$(2): \quad \langle f_n\rangle$ converges uniformly to $f$


Then:

$f$ is continuous at every point of $M$.


Proof

Let $a \in M$.

We are given that $d_N$ is a metric on $N$.

So:

$\forall x, y, z \in N: d_N \left({x, z}\right) \le d_N \left({x, y}\right) + d_N \left({y, z}\right)$

We apply this property twice to assert that $\forall n \in \N, \forall x \in M$, we have:

\(\text {(1)}: \quad\) \(\ds d_N \left({f \left({x}\right), f \left({a}\right)}\right)\) \(\le\) \(\ds d_N \left({f \left({x}\right), f_n \left({x}\right)}\right)\)
\(\ds \) \(+\) \(\ds d_N \left({f_n \left({x}\right), f_n \left({a}\right)}\right)\)
\(\ds \) \(+\) \(\ds d_N \left({f_n \left({a}\right), f \left({a}\right)}\right)\)


Let $\epsilon >0$.

Since $f_n \to f$ uniformly:

$\exists \mathcal N \in \R: \forall n \ge \mathcal N: \forall x \in M$:
\(\text {(2 a)}: \quad\) \(\ds d_N \left({f \left({a}\right), f_n \left({a}\right)}\right)\) \(<\) \(\ds \frac \epsilon 3\)
\(\text {(2 b)}: \quad\) \(\ds d_N \left({f \left({x}\right), f_n \left({x}\right)}\right)\) \(<\) \(\ds \frac \epsilon 3\)


We are given that $\forall n \in \N: f_n$ is continuous.

So, for any $n\in\N$, $\exists \delta > 0: \forall x \in M$:

\(\text {(3)}: \quad\) \(\ds d_M \left({x, a}\right)\) \(<\) \(\ds \delta\)
\(\ds \implies \ \ \) \(\ds d_N \left({f_n \left({x}\right), f_n \left({a}\right)}\right)\) \(<\) \(\ds \frac \epsilon 3\)


By combining $(1)$, $(2 a)$, $(2 b)$ and $(3)$:

$\exists \delta > 0$ and $\exists n$ sufficiently large such that $\forall x \in M$:
\(\ds d_M \left({x, a}\right)\) \(<\) \(\ds \delta\)
\(\ds \implies \ \ \) \(\ds d_N \left({f \left({x}\right), f \left({a}\right)}\right)\) \(\le\) \(\ds \frac \epsilon 3 + \frac \epsilon 3 + \frac \epsilon 3\)
\(\ds \) \(=\) \(\ds \epsilon\)


As $a$ and $\epsilon$ are arbitrary, it follows that:

$\forall a \in M: \forall \epsilon > 0: \exists \delta > 0: \forall x \in M$:
\(\ds d_M \left({x, a}\right)\) \(<\) \(\ds \delta\)
\(\ds \implies \ \ \) \(\ds d_N \left({f \left({x}\right), f \left({a}\right)}\right)\) \(\le\) \(\ds \epsilon\)


Hence, $f$ is continuous at every point of $M$.

$\blacksquare$


Sources