On which subspace $W\subset C^{\infty}[0,1]$ is $D(f)=xf'(x)$ a bounded operator provided all functions in $W$ are flat functions? First we introduce the concept &quot;Flat function&quot;:
Definition: A smooth function $f:\mathbb{R}\to \mathbb{R}$ is called a flat function at origin if $f^{(k)}(0)=0$ for all $k=0,1,2,\ldots $.
let $V=\{f\in C^{\infty}[0,1]\mid \text{f is flat at origin}\}$. We equip $V$ with $|\quad|_{\infty}$.
Is there an infinit dimensional subspace $W\subseteq V$ which is invariant under the differential operator
$D(f)=xf'(x)$ and $D$ is a bounded operator on $W$?
The motivation comes from page 43 remark 2 of the following paper:
http://mcs.qut.ac.ir/article_243944.html
To me, it seems unlikely that such a subspace $W$ can exist. If you define $Q:V \to V$ by $(Qf)(x) = \log(x) f(x)$, then you can check that $$QD-DQ=\mathrm{id}_V$$ which is a version of the canonical commutation relation. I don't know off the top of my head whether realizing the canonical commutation relation while having one of operators bounded is out of the question in the setup you have given, but it at least seems rather far fetched.
Added: Actually, what I wrote above may not be an issue after all. What if $Q$ doesn't map  $W$ (or a dense subspace of $W$) back into $W$? I'll leave this answer up anyway just in case it is somehow helpful.

Other thoughts:
Let me also throw in some disorganized  notes which might be helpful (and which are anyway the way I arrived at the operator $Q$ above).

$D=x \frac{d}{dx}$ is the generator of the scaling action $(\phi_t f)(x)=f(e^tx)$. Actually, if your domain is in $C^\infty([0,1])$, it is more natural to reverse the flow because we only have a sensible scaling map $\phi_t : C^\infty([0,1]) \to C^\infty([0,1])$ when $t \leq 0$ and you would probably prefer a 1-parameter semigroup defined for $t \geq 0$. Boundedness of $x \frac{d}{dx}$ would correspond to norm-continuity (as opposed to strong operator continuity) of this semigroup on the subspace $W$.
You can check that the map $\iota:C^\infty((0,1]) \to C^\infty([1,\infty))$ defined by $(\iota f)(x)=f(1/x)$ maps your space $V \subseteq C^\infty((0,1])$ exactly onto the subspace $V' \subseteq C^\infty([1,\infty))$ consisting of functions all of whose derivatives decay rapidly at $+\infty$. That is the Schwartz functions, restricted to $[1,\infty)$.
If you conjugate the scaling action through the inversion map $\iota$ above, it is still the same scaling action (though it gets time reversed).
Now you can use another identification  $j : C^\infty([1,\infty) \to C^\infty([0,\infty))$ defined by $(j f)(x)=f(e^x)$. This carries $V'$ onto a subspace $V''$ which is contained in the subspace of functions in $C^\infty([0,\infty))$ with Schwartz behaviour at infinity. You don't hit them all though, for example $x \mapsto e^{-x}$ is a Schwartz function in $C^\infty([0,\infty))$ which does not belong to $V''$.
Under the identification $j$, the scaling action becomes the translation flow  on $C^\infty([0,\infty))$. So the operator $x \frac{d}{dx}$ is then exchanged for the usual differentiation operator $\frac{d}{dx}$.
You could try furthermore taking a Fourier transform at this point to get some subspace of the Hardy space of the line, and replace the operator with multiplication by $x$, but this seems a bit pointless because it is not very clear how to deal with the uniform norm under the Fourier transform.

Anyway, one upshot of these observations is that, if there does exist a subspace $W$ with the property you are after, then there will also exist an infinite-dimensional subspace $W'' \subset C^\infty([0,\infty))$ satisfying

$W''$ is invariant under translation (towards $0$).
The elements of $W''$ have Schwartz function behaviour at $\infty$.
$\frac{d}{dx}$ acts boundedly on $W''$ (with respect to the uniform norm).
Equivalently to (3), the translation semi-group is norm-continuous on $W''$. I guess that forces some kind of uniformly equicontinuity condition on $W''$, since the uniform norm is being used.
Let $W=&lt;f_1,f_2,...,&gt;$ then let $f=\sum_i c_i f_i$. $D(f) = \sum_i c_i D(f_i)$.
Take $f_i(x) = x^i$. This $W$ is flat and invariant under $D$ and infinite dimensional but if u wanted boundedness of $D$ then $deg(f_i)$ might cause issue but u might just be able to prove boundedness since $x^n &lt; x^{n-1}$ for $0 \leq x &lt; 1$ i.e., if u remove a small interval near $1$ from ur domain or norm.
On the other hand:
Let us apply $D$ several times. $D(D(f)) = D(xf^{(1)}(x)) = x^2 f^{(2)}(x)+xf^{(1)}(x)$. $D^3(f)=D(x^2 f^{(2)}(x)+xf^{(1)}(x)) = x^3 f^{(3)}(x)+3x^2f^{(2)}(x)+xf^{(1)}(x)$. You can see that the coefficient of $x^n f^{(n)}(x)$ keeps growing.
YOu can see that coefficient of $x^n f^{(n)}(x)$ is $O(n!)$ whereas we can only handle growth by $||D||^n$. This says how fast $f^{(n)}(x)$ must decay. Does this help ?