About density of some subsets of infinitely differentiable functions in $C[0,1]$

Question

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?

Answer 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 $P_{k_1,\dotsc, k_{i-1}}$.

Answer 2

Yes, it is dense.

Given $f \in C([0,1])$ and $\epsilon > 0$, it is easy to find a continuous function $g$ which is constant on some $\delta$-neighborhood of each $x_i$ and has $\|f-g\| < \epsilon$. (For instance, Now extend $g$ continuously to all of $\mathbb{R}$, say by making it constant on $(-\infty, 0]$ and $[1,\infty)$. Let $\phi$ be a $C^\infty$ bump function with $\int \phi = 1$ and compactly supported inside $(-\delta, \delta)$. Then the convolution $h = g \ast \phi$ is also constant on some neighborhood of each $x_i$, so $h^{(k)}(x_i) = 0$ for all $k \ge 1$, and $h \in C^\infty$, so $h$ is in your set. By choosing the support of $\phi$ as small as needed and using the uniform continuity of $g$, we can get $\|g-h\| < \epsilon$.