# Making dense subset of continuous functions using a single continuous function

I have a question about set of continuous function on the compact interval.

Denote the set of all continuous $$n$$-dimensional real functions on $$[0,L]$$ as $$\mathcal{C}_{[0,L]}$$

($$\mathcal{C}_{[0,L]} = \left\{ u: [0,L] \rightarrow \mathbb{R}^n\right\}$$)

Is there any $$T>L$$ and a continuous function $$f: [0,T] \rightarrow \mathbb{R}^n$$ such that

finite combination of $$L$$-length segments of $$f$$ is dense in $$\mathcal{C}_{[0,L]}$$ with respect to supremum norm?

I mean if we define $$f_{[t,t+L]}:[0,L] \rightarrow \mathbb{R}^n$$ as $$f_{[t,t+L]}(\cdot) = f(\cdot + t)$$ for $$t \in [0,T-L]$$, is it possible to find continuous function $$f: [0,T] \rightarrow \mathbb{R}^n$$ such that any finite linear combination of $$f_{[t, t+L]}$$, is dense subset of $$\mathcal{C}_{[0,L]}$$ with respect to sup-norm.

If not, is there any space of functions that satisfies a similar property? (e.g., rather $$L^2$$ space satisfies such a condition etc.)

• We recently discussed something similar but in $L^2$. In that space the problem is easier, since there is the Tauberian theorem of Wiener which completely answers the question: math.stackexchange.com/a/4441449/8157 Commented Jan 13, 2023 at 10:41
• The following is the Tauberian theorem of Wiener: en.wikipedia.org/wiki/Wiener%27s_Tauberian_theorem Commented Jan 13, 2023 at 10:42
• Thanks for a comment. It seems Problem of Wiener Tauberian theorem little different from my problem because it is considering translates but I'm considering varying time windows.
– 박희인
Commented Jan 16, 2023 at 3:58

The answer is positive for $$n=1.$$
Let $$L=2\pi$$ and $$T=4\pi.$$ For a fixed $$0 for example $$r={1\over 2},$$ let $$f(x)=1+2\sum_{n=1 }^\infty r^{n}\cos nx ={1-r^2\over 1-2r\cos x+r^2}$$ Then for $$0\le t\le 2\pi$$ we get $$f_t(x)=1+2\sum_{n=1 }^\infty [r^{n}\cos nt\cos nx -r^{n}\sin nt\sin nx]$$ Assume, by contradiction, that the linear span, with real coefficients, of the functions $$f_t$$ is not dense in $$C_{\mathbb{R}}[0,2\pi].$$ Then there exists a nontrivial bounded linear functional $$\varphi$$ on $$C_{\mathbb{R}}[0,2\pi]$$ vanishing on every $$f_t.$$ Hence $$\varphi(f_t)=\varphi(1)+2\sum_{n=1}^\infty [r^n\varphi(\cos nx)\cos nt-r^n\varphi(\sin nx)\sin nt]=0 \quad 0\le t\le 2\pi$$ Thus the Fourier coefficients of the function $$[0,2\pi]\ni t\mapsto\varphi(f_t)$$ vanish, hence $$\varphi(1)=0,$$ $$\varphi(\cos nx)=$$ and $$\varphi(\sin nx)=0.$$ As trigonometric polynomials are dense in $$C_\mathbb{R}[0,2\pi]$$ we get $$\varphi=0,$$ a contradiction.