Some time ago, I asked about inite interpolation by
a nondecreasing polynomial here at [Finite interpolation by a nondecreasing polynomial](https://mathoverflow.net/questions/16673/finite-interpolation-by-a-nondecreasing-polynomial). This turned out to be an already solved problem; it also turned out that the degree of the solution could not be bounded in terms of the number of interpolation points alone.


My new question is: if we are willing to replace polynomials by another
vector space V of indefinitely differentiable functions, then we can we achieve something better than with polynomials, in the sense that V is finite-dimensional?


Formally, fix $x\_1 \lt x\_2 \lt \ldots \lt x\_n$ and let $y\_1 \leq y\_2\leq \ldots \leq y\_n$ vary. We consider the system $(S)$ made of the $n$ interpolation constraints $f(x\_i)=y\_i$ for $i$ between $1$ and $n$. Is there a finite-dimensional subspace $V$ of ${\cal C}^{\infty}([x\_1,x\_n],{\mathbb R})$ such that for any $y\_1 \leq y\_2\leq \ldots \leq y\_n$, there is a solution to $(S)$ which is nondecreasing and also in $V$?



Finite interpolation by nondecreasing indefinitely differentiable functions in a finite-dimensional space
This is to expand Qiaochu Yuan's comment. For $1\le i < n$ let $b\_i:=(x\_{i+1}+x\_i)/2$ be the mid-point of the $i$-th interval, and let $0 < \epsilon \leq \min\_{1\le i < n} (x\_{i+1}+x\_i)/2\, .$ Start with the linear space $V\_0$ of all continuous functions on $\mathbb{R}$ that are affine on each component interval of $\mathbb{R}\setminus \{b \_i\, : \, 1\leq i < n \}$ and take $V$ to be the image of $V\_0$ via the (linear) convolution operator $u\mapsto u\*\phi$, with a symmetric kernel $\phi\ge0$, $C^\infty\_c(\, ]-\epsilon,+\epsilon[\, )\, $ and $\int\_\mathbb{R} \phi = 1 .$ The convolution with $\phi$ preserve monotonicity and the values at the nodes $x\_i$, so that the interpolation function in $V$ is just the mollification of the interpolation function on $V\_0.$


