Consider the operator \begin{eqnarray*} K_{n} &:&L^{2}(0,1)\longrightarrow L^{2}(0,1)^{n}, \\ u(x) &\mapsto &A_{n}U_{n}(x)=A_{n}(u(\frac{x}{n}),u(\frac{x+1}{n}),...,u(% \frac{x+n-1}{n}))^{t} \end{eqnarray*}
where $A_{n}$ is $n\times n$ matrix with $\det (A_n)\neq 0,$ $\forall n\geq 1.$
It is clear that $K_{n}$ is onto since for any $Y\in L^{2}(0,1)^{n}$ the equation $A_nU_n(x)=Y(x)$ admits a unique solution in $L^{2}(0,1)^{n}$which makes $u$ uniquely determined in this case.
I'm wondering about the surjectivity property for $n\mapsto \infty .$ If we assume that $\underset{n\longrightarrow \infty }{\lim }\det (A_n)\neq 0$ do we obtain that the limit operator $K_{\infty }$ is surjective?.
I was thinking of studying the adjoint $K_{n}^{\ast \text{ }}$ and proving that $K_{n}^{\ast \text{ }}v_{n}=0$ for some $\left\Vert v_{n}\right\Vert _{L^{2}(0,1)^{n}}=1$ but I didn't succeed. Any ideas?. Thank you in advance?
