Stability of the uniform convergence under division by the norm of the argument

Question

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.

Answer 1

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 $f=O(||x||)$

Answer 2

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.