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I'm thinking about the following problem and get stuck.

I consider a complete orthonormal set $\{\psi_i\}^{\infty}_{i=0}$ defined on $\mathbb{R}$, and the inner product is defined by $$(\psi_i,\psi_j)=\int^{\infty}_{-\infty}\psi_i(x)\psi_j(x)\,dx$$ In this sense, I consider a finite-dimensional space $E_k$ spanned by $\{\psi_i\}^{k-1}_{i=0}$, and assume that $\psi_k$ has $(k+1)$ zeros, and denote these $(k+1)$ zeros of $\psi_k$ by $\beta_1,\beta_2,\cdots,\beta_{k+1}$. Now I define a space $\Phi_{k+2}$ spanned by $$ \begin{aligned} &\phi_1(x)= \begin{cases} \psi_k(x)\quad \text{for}\ x\in(-\infty,\beta_1),\\ 0\quad \text{otherwise} \end{cases}\\ &\phi_2(x)= \begin{cases} \psi_k(x)\quad \text{for}\ x\in(\beta_1,\beta_2),\\ 0\quad \text{otherwise} \end{cases}\\ &\cdots\\ &\phi_{k+2}(x)= \begin{cases} \psi_k(x)\quad \text{for}\ x\in(\beta_{k+1},\infty),\\ 0\quad \text{otherwise} \end{cases} \end{aligned} $$ Obviously, $\{\phi_i\}^{k+2}_{i=1}$ is linearly independent, so it's a basis. There is a claim that $\dim({E^\perp_k\,\cap\, \Phi_{k+2}})\geq2$. What I know is at least, $\phi_k\in{E^\perp_k\,\cap\, \Phi_{k+2}}$, but where is the other basis? Appreciate for any help.

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    $\begingroup$ $\Phi_{k+2}$ is a $k+2$ dimensional space. $E_k$ is a $k$ dimensional space. Even if $E_k \subset \Phi_{k+2}$, there wouls still have to be two dimensions of $\Phi_{k+2}$ perpendicular to $E_k$. $\endgroup$ Commented Oct 18, 2023 at 20:02
  • $\begingroup$ @PaulSinclair Awesome! $\endgroup$
    – xfireskyx
    Commented Oct 28, 2023 at 0:28

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