Uniform convergence of $f_n$ on a compact implies uniform convergence on some $\delta$-enlargement of this compact We have $(D,d)$ a metric space, $K$ a compact of D.
$f_n$ is a decreasing sequence of positive real-valued continuous bounded functions on $D$ that converges uniformly to $0$ on $K$ and pointwise to $0$ on $D$. We define $K^\delta = \{ x \in D : d(x,K) < \delta \}$. How can one show that there exists a $\delta > 0$ such that $f_n$ converges uniformly to $0$ on $K^\delta$. I feel like we should use uniform continuity on $K$, but I don't see where.
 A: This isn't true without further assumptions on $D$.
For example, it's true if $D$ is locally compact, as clark already alluded to.  Indeed, if $x \in K$, then there is a $\delta_{x} > 0$ such that $B(x,\delta)$ is pre-compact.  Then, by Dini's Theorem, $(f_{n})_{n \in \mathbb{N}}$ converges uniformly on $\overline{B(x,\delta)}$.  By compactness, we can cover $K$ by finitely many of these open balls.  Call their union $U$.  Appealing to compactness again, there is a $\delta > 0$ such that $K \subseteq K^{\delta} \subseteq U$.  By construction, $(f_{n})_{n \in \mathbb{N}}$ converges uniformly on $U$ so it converges uniformly on $K^{\delta}$.  
To see this isn't true in general, take $D = \ell^{1}(\mathbb{N})$ with the metric induced by the norm.  For each $n \in \mathbb{N}$, define a linear functional $F_{n} : \ell^{1}(\mathbb{N}) \to \ell^{1}(\mathbb{N})$ by
$$F_{n}(x_{1},x_{2},\dots) = \left(\frac{x_{1}}{n}, \frac{x_{2}}{n^{\frac{1}{2}}}, \dots, \frac{x_{j}}{n^{\frac{1}{j}}},\dots\right).$$  Observe that $\|F_{n}\| =1$.  Thus, $(F_{n})_{n \in \mathbb{N}}$ are continuous functions.  Finally, define $f_{n} : \ell^{1}(\mathbb{N}) \to \mathbb{R}_{\geq 0}$ by
$$f_{n}(x_{1},x_{2},\dots) = \sum_{j = 1}^{\infty} \frac{|x_{j}|}{n^{\frac{1}{j}}} = \|F_{n}(x_{1},x_{2},\dots)\|_{\ell^{1}(\mathbb{N})}.$$
$(f_{n})_{n \in \mathbb{N}}$ is a sequence of continuous functions on $\ell^{1}(\mathbb{N})$ converging pointwise to zero.  Moreover, since the sum defining $f_{n}$ decreases termwise, it follows that $f_{n + 1} \leq f_{n}$.  Thus, $(f_{n})_{n \in \mathbb{N}}$ is a decreasing sequence of non-negative continuous functions that converges pointwise to zero on $\ell^{1}(\mathbb{N})$, which is what we're looking for.  
(For my fellow pedants, if we insist on a sequence of positive functions, we can instead use $$g_{n}(x_{1},x_{2},\dots) = f_{n}(x_{1},x_{2},\dots) + n^{-1},$$ which is a decreasing sequence of positive continuous functions.  The arguments that follow work for $(g_{n})_{n \in \mathbb{N}}$ just as much as they work for $(f_{n})_{n \in \mathbb{N}}$.)
Now let $K = \{0\}$.  I claim that if $\delta > 0$, then $(f_{n})_{n \in \mathbb{N}}$ does not converge uniformly on $B(0,\delta)$.  To see this, fix $\delta > 0$ and let $\{e_{j}\}_{j \in \mathbb{N}}$ denote the sequences 
$$e_{1} = (1,0,0,\dots), \, \, e_{2} = (0,1,0,\dots), \dots$$
Observe that if $n, j \in \mathbb{N}$ and $\lambda \in (0,1)$, then $\lambda \delta e_{j} \in B(0,\delta)$ and
$$f_{n}(\lambda \delta e_{j}) = \lambda \delta n^{-\frac{1}{j}}.$$
By taking $j \to \infty$ and $\lambda \to 1$, we obtain
$$\sup \{f_{n}(x) \, \mid \, \ x \in B(0,\delta)\} = \delta.$$
Thus, $(f_{n})_{n \in \mathbb{N}}$ does not converge to zero uniformly on $B(0,\delta)$.  Since $K^{\delta} = B(0,\delta)$, this concludes the proof.  
