Uniform Convergence And Continuity

Uniform Convergence Equivalence In Terms of Lim Sup

Suppose that

S \subseteq ℝ^{n}

and a sequence of functions

(f_{k}) : ℕ_{1} \to C (S, ℝ^{m})

, then

f_{k} \to f

if and only if eventualy we have

f_{k} - f \in C_{b} (S, R^{m})

for all

k

sufficiently large and

\lim_{k \to \infty} {‖ f_{k} - f ‖}_{\infty} = 0

Dini's

f

and

f_{n}

are continuous functions on

[a, b]

such that

f_{n} \geq f_{n + 1}

for all

n \in ℕ_{1}

and if

f_{n} \overset{pw}{\to} f

then

f_{n} \overset{unif}{\to} f

We first prove that for any $x_{0} \in [a, b]$ and $ϵ \in ℝ^{+}$ there exists some integer $N \in ℕ_{1}$ and $r \in ℝ^{+}$ such that $| x - x_{0} | \leq r ⟹ | g_{N} (x) | \leq ϵ$ where $g_{n} (x) = f (x) - f_{n} (x)$

This follows from a few facts, we first start with continuity of $f, f_{n}$ , specifically at $x_{0}$ and therefore using $ϵ$ we obtain some $δ_{1} \in ℝ^{+}$ such that $f (B (x_{0}, δ_{1})) \subseteq B (g_{n} (x_{0}), ϵ)$ , similarly we have a $δ_{2} \in ℝ^{+}$ such that $f (B (x_{0}, δ_{2})) \subseteq B (f (x_{0}), ϵ)$ , then also we know that since $f_{n} \overset{pw}{\to} f$ then since it's true at $x_{0}$ and using $ϵ$ we obtain some $K \in ℕ_{1}$ such that for all $n \in ℕ_{K}$ we have $| f_{n} (x_{0}) | - f (x_{0}) \leq ϵ$

Now we combine using the triangle inequality, so let $ϵ \in ℝ^{+}$ , then let $r = \min (δ_{1}, δ_{2})$ so that we obtain the inequalities mentioned in the previous paragraph, and we get some $K \in ℕ_{1}$ as mentioned in the previous paragrah, then $\begin{matrix} | g_{K} (x) | & = | f (x) - f_{K} (x) | \\ \leq | f (x) - f (x_{0}) | + | f (x_{0}) - f_{K} (x_{0}) | + | f_{K} (x_{0}) - f_{K} (x) | \\ \leq 3 ϵ \end{matrix}$

With that out of the way suppose for the sake of contradiction that $f_{n}$ doesn't convege uniformly to $f$ , since $f - f_{n}$ is eventually bounded then this is equivalent to $\lim_{n \to \infty} {‖ f - f_{n} ‖}_{\infty} = d \in ℝ^{+}$

Since $g_{n}$ is continuous as the sum of continuous functions and has compact domain then by EVT, we know that it attains its max and min, and therefore there exists a $x_{k} \in [a, b]$ such that ${‖ f - f_{k} ‖}_{\infty} = g_{k} (x_{k})$ , so we have $\lim_{k \to \infty} g_{k} (x_{k}) = d$ , and note that this means that for any $ϵ^{'} \in ℝ^{+}$ there is some $J \in ℕ_{1}$ such that for all $j \in ℕ_{J}$ we have $| g_{j} (x_{j}) - d | < ϵ^{'}$ which implies that $d - ϵ^{'} \leq g_{j} (x_{j})$ .

Since $x_{k}$ is in a compact set it has a convergent subsequence $x_{n_{k}} \to x \in [a, b]$ . Now recall the fact we proved initially, by using $d - ϵ$ we know that there is some $N \in ℕ_{1}$ such that $g_{N} (x) < d - ϵ$ , now at the same time the inequality discussed in the previous paragraph also holds on the subsequence (this is because if a sequence goes somewhere, then any subsequence must also go there), namely we have some $J \in ℕ_{1}$ such that for all $j \in ℕ_{J}$ $d - ϵ \leq g_{n_{j}} (x_{n_{j}})$ but then there is some $i$ such that $n_{i} > \max (N, J)$ and therefore $g_{i} < g_{N}$ and so chaining inequalities we have $d - ϵ \leq g_{n_{i}} (x_{x_{i}}) \leq g_{N} (x) \leq d - ϵ$ which is a contradiction, and therefore $f_{n} \overset{unif}{\to} f$