Making dense subset of continuous functions using a single continuous function

Question

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.)

Thank you for your attention.

Answer 1

The answer is positive for $n=1. $

Let $L=2\pi$ and $T=4\pi.$ For a fixed $0<r<1,$ 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.