Stability of the uniform convergence under division by the norm of the argument Let $U$ be an open subset of the normed vector space $V$.
Let $f_n$ be a sequence of continuous functions with signature $U \to 
\mathbb{R}$ converging uniformly to $\varphi$.
Assume that all functions has the property that
$$
\lim_{x \to 0} \frac{f_n(x)}{\|x\|} = 0
$$
which we denote as $f_n(x) = O(\|x\|)$.
In case sequence $\frac{f_n(x)}{\|x\|}$ converges uniformly over some punctured neighborhood of 0 then $\varphi = O(\|x\|)$ by the the virtue of Moore-Osgood theorem.
However I don't know if convergence of quotients can be inferred directly from the properties of the sequence:

Is it true, that the set $\{ f \in C_
\infty(U) : f(x) = O(\| x \|) \}$ is a closed subspace of $C_\infty(U)$,  the Banach$\,^\dagger$ space of continuous functions with sup-norm?

Furthermore, I am curious if in this case sequence $\frac{f_n(x)}{\|x\|}$ will be uniformly convergent itself?
($\dagger$) infinite values of the norm are possible in this space as $U$ is not compact , but I don't think that this is important in the context of the question.
 A: Let $f_n \in A=\{ f \in C_
\infty(U) : f(x) = O(\| x \|) \}$ such that $f_n \rightarrow f$ uniformly.
Let $\epsilon>0$.
Then if $ x \neq 0$,
we have that exists $n_0 \in \mathbb{N}$ such that  $||f_n-f||_{\infty}\leqslant \epsilon||x|| ,\forall n \geqslant n_0,\forall x \in U$
Also $f_n=O(||x||), \forall n \in \mathbb{N} $
Thus $$ \frac{|f(x)|}{||x||} \leqslant \frac{|f_{n_0}(x)-f(x)|}{||x||}+\frac{|f_{n_0}(x)|}{||x||} \leqslant \epsilon + \frac{|f_{n_0}(x)|}{||x||}$$
Then $\limsup_{x \rightarrow 0}\frac{|f(x)|}{||x||} \leqslant \epsilon +0=\epsilon$ thus $f=O(||x||)$
A: With the help of Marios Gretsas I will try to answer last question.
We know that $f_n \rightarrow \varphi$ implies $\varphi(x) = O(\|x \|) $.
Fix $\varepsilon > 0$.
By this result it is possible to select such $\delta > 0$ that for all $x \in U$ such that $\| x \| \le \delta$ we have 
$$
\frac{|\varphi(x)|}{\|x\|} < \frac{\varepsilon}{4}.
$$
This means that there is such an integer $N_1$ that for all $n \ge N_1$ and same $x$,
$$
\frac{|f_n(x)|}{\|x\|} < \frac{\varepsilon}{2}. 
$$
From uniform convergence it follows that there is integer $N_2$ such that for all $n,m \ge N_2 $ we have
$$
 \sup_{x \in U} | f_n(x) - f_m(x) |  < \delta\varepsilon. 
$$
Select $N = \max \{ N_1, N_2 \} $ and fix $n,m \ge N$. Let $B = \{ u \in U : \| u\| \le \delta  \} $.  Then,
$$
 v_1 = \sup_{x \in B} \left| \frac{f_n(x)}{\|x\|} - \frac{f_m(x)}{\|x\|}  \right| \le  \sup_{x \in B} \frac{| f_n(x) |}{\|x\|} + \frac{|f_m(x)|}{\|x\|} < \varepsilon,
$$ 
$$
v_2 = \sup_{x \in U \setminus B}  \frac{|f_n(x) - f_m(x)|}{\|x\|} \le \sup_{x \in U\setminus B} \frac{\delta \varepsilon}{\|x\|} < \varepsilon.
$$
Clearly,
$$
 \sup_{x \in U} \frac{| f_n(x) - f_m(x)|}{\|x\|} = \max\{v_1,v_2\} < \varepsilon,
$$
hence, $\frac{f_n(x)}{\|x\|}$ is Cauchy in sup-norm and, thus, it must converge uniformly.
Edit 
It seems that the first statement works only if $f_n$ is equicontinuous with strictly sublinear modulus of continuity in some neighborhood of $0$. 
Consider counterexample:
decreasing saw-tooth functions
$$ 
f_n(x) =
\begin{cases}
 2x - 2^{-n} \quad \textrm{if} \quad x \in \Big[2^{-n - 1},2^{-n}\Big) \\
 2^{-n} + 2^{-n + 1}  - 2x \quad \textrm{if} \quad x \in \Big[ 2^{-n}, 2^{-n} + 2^{-n - 1}\Big)\\
0 \quad \textrm{otherwise}
\end{cases}
$$
then $\| f_n \| = 2^{-n} \to 0$, hence $f_n \to 0$. 
But
$$
\sup_{x \in \mathbb{R} \setminus \{0\}  }
\left\| 
   \frac{f_n(x)}{\|x\|}
\right\| \ge 1
$$
so this construction does not converge to $0$. $f_n(x) = O(\|x\|)$ as they all are zero the neighborhood of 0.
