Let $x\_1,...,x\_m$ be fixed numbers from $[0,1]$ and let $k\_1,..., k\_m$ be fixed natural numbers ($\geq 1$).
Is the set
$$\{f\in C^\infty[0,1]: f^{(k\_1)}(x\_1)=0,...,f^{(k\_m)}(x\_m)=0 \}$$a dense subset of the Banach space $C[0,1])$, with the supremum norm?



About density of some subsets of infinitely differentiable functions in $C[0,1]$
Here is a solution. We prove by induction on $m$. Denote by $P$ the subspace of $C([0,1])$ consisting of the restrictions to $[0,1]$ of the smooth functions $\mathbb{R}\to\mathbb{R}$. Assume first that $x\_1,\dotsc, x\_m$ are pairwise distinct. Set


$$ P\_{x\_1,\dotsc, x\_k}:=\big\{ p\in P;\;\;p^{(k\_i)}(x\_i)=0,\;i=1,\dotsc, m\big\}. $$


Then


$$ P\supset P\_{x\_1},\supset \cdots \supset P\_{x\_1,\dotsc, x\_m}. $$


Let us show that $P\_{x\_1,\dotsc, x\_m}$ is dense in $P$. The case $m=1$ is dealt with as in my comment.


Let us now prove that $P\_{x\_1,\dotsc, x\_m}$ is dense in $P\_{x\_1,\dotsc, x\_{m-1}}$, thus, inductively, in $P$.


Define the linear functional $\newcommand{\bR}{\mathbb{R}}$


$$ L: P\_{x\_1,\dotsc, x\_{m-1}}\to \bR,\;\;L(p)= p^{(k\_m)}(x\_m). $$


The linear functional $L$ is not continuous thus its kernel $\ker L=P\_{x\_1,\dotsc, x\_m}$ is dense in $P\_{x\_1,\dotsc, x\_{m-1}}$.


To see that $L$ is not continuous pick a smooth function $f$ with compact support contained in a neighborhood of $x\_m$ disjoint form $x\_1,\dotsc, x\_{m-1}$ and such that $f^{(k\_m)}(x\_m)=1$. Define


$$f\_n(x)= n^{-1/2}f\big( x\_m+ n(x-x\_m)\big).$$


Then the sequence $f\_n$ converges uniformly to zero but $$f\_n^{(k\_m)}(x\_m)=n^{k\_m-1/2}\to\infty.$$


Suppose now that $x\_1,\dotsc, x\_m$ are not necessarily distinct. For simplicity assume that $x\_1=\cdots=x\_m$ and $k\_1<\cdots <k\_m$. Then define


$$ P\_{k\_1,\dotsc, k\_m}:=\big\{ p\in P;\;\;p^{(k\_i)}(x\_1=0,\;\;i=1,\dotsc, m\big\}. $$
Arguing as above one sees that $P\_{k\_1,\dotsc, k\_i}$ is dense in $P\_{k\_1,\dotsc, k\_{i-1}}$.


