$\dim\{f:(-1,1)\to \mathbb{R}\mid f^{(n)}(0)=0 ~\forall~ n~\geq0\}=\infty$ if $f\in C^\infty[-1,1]$

Claim: $$V=\{f:(-1,1)\to \mathbb{R}\mid f^{(n)}(0)=0 ~\forall~ n~\geq0\}$$ is infinite dimensional over $\mathbb{R}$ where $f$ is infinitely differentiable on $(-1,1)$.

Attempt: Before handling the dimension I'm wondering what functions does $V$ contain. If we remove the property $f^{(n)}(0)=0 ~\forall~ n~\geq0$ of the functions of $V$ we get that $V$ contains polynomials, exponential functions, trigonometric functions but non of these have $f^{(n)}(0)=0 ~\forall~ n~\geq0$. So, all these types of functions that I mentioned won't work except for the zero function but that makes the dimension of $V$ $1$ if I consider that $0$ is the only possible function in $V$ which is not the case. What are those functions that I'm missing? Any hints?

Solution 1:

Hint: $e^{-n/x^{2}}, n=1,2,...$ are linearly independent elements of $V$. (These functions are given the value $0$ at $x=0$).

It helps to make the change of variable $y=\frac 1 x$ in the proof of independence.