# dimension of orthogonal complement of complete orthonormal sets

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.

• $\Phi_{k+2}$ 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 $E_k$. Commented Oct 18, 2023 at 20:02
• @PaulSinclair Awesome! Commented Oct 28, 2023 at 0:28