Suppose $\{e_i\}_{i\in I}$ is an orthonormal basis for some Hilbert space. $\{f_i\}_{i\in I}$ is an orthonormal family with the property $\sum^\infty_{i\in I} \|e_i-f_i\|^2<\infty$, show that $\{f_i\}_{i\in I}$ is also orthonormal basis.

I can think of for a $x$ with $(x,f_i)=0$ for all $i$, prove $x=0$. However, I don't know how to continue to the next step.

  • 1
    $\begingroup$ Are the bases countable? $\endgroup$
    – copper.hat
    Apr 29 '17 at 5:36
  • $\begingroup$ It may not be countable. $\endgroup$
    – user284873
    Apr 29 '17 at 5:46
  • 2
    $\begingroup$ I am pretty sure that this is problem 12 in "A hilbert space problem book" by Halmos. But I currently don't have the book with me. $\endgroup$
    – PhoemueX
    Apr 29 '17 at 6:12
  • $\begingroup$ I think it is fairly straightforward to show that such an $x$ must be in the span of a finite number of $e_i$, but the remainder of the proof escapes me now. $\endgroup$
    – copper.hat
    Apr 29 '17 at 6:14
  • 1
    $\begingroup$ @Kevin That first solution looks surprisingly elementary. The second solution seems to be a variation on my answer below (or, perhaps more appropriately, my answer seems to be a variation on the second solution). $\endgroup$ May 24 '17 at 8:39

Let $H$ denote the given Hilbert space. As noted by copper.hat, one may assume $I=\mathbb{N}$. First observe that $$ \sum_{i=1}^\infty \lvert 1 - \langle e_i,f_i\rangle\rvert^2 + 2\sum_{i=1}^\infty \sum_{j=i+1}^\infty \lvert \langle f_j,e_i\rangle\rvert^2 = \sum _{i = N + 1} ^{\infty} \lVert e _i - \sum_{j=1}^\infty \langle e_j,f_i\rangle e_j \rVert ^{2} = \sum _{i = 1} ^{\infty} \lVert e _i - f _i \rVert ^{2}. $$ Pick $N\in\mathbb{N}$ such that \begin{align*} \sum _{i = N + 1} ^{\infty} \lVert e _i - f _i \rVert ^{2} &< 1/2, \\ \sum_{i=1}^{N} \sum_{j=N+1}^\infty \lvert \langle f_j,e_i\rangle\rvert^2 &< 1/2. \end{align*}

Define $E_N=\overline{\mbox{span}}(e_{N+1},e_{N+2},\ldots)$ and $F_N=\overline{\mbox{span}}(f_{N+1},f_{N+2},\ldots)$. For a closed subspace $A\subseteq H$, let $P_A$ denote the orthogonal projection onto $A$ in $H$. Then we find \begin{align*} \lVert P_{E_N} - P_{F_N} \rVert_{\text{HS}}^2 & = \sum_{i=1}^N \lVert P_{F_N}e_i \rVert^2 + \sum_{i=N+1}^\infty \lVert e_i - P_{F_N}e_i \rVert^2 \\ & = \sum_{i=1}^N \sum_{j=N+1}^\infty \lvert\langle f_j,e_i\rangle \rvert^2 + \sum_{i=N+1}^\infty \lVert e_i - P_{F_N}e_i \rVert^2 \\ & \leqslant \sum_{i=1}^N \sum_{j=N+1}^\infty \lvert\langle f_j,e_i\rangle \rvert^2 + \sum_{i=N+1}^\infty \lVert e_i - f_i \rVert^2 < 1. \end{align*} Thus, we have $$ \lVert P_{E_N^\perp} - P_{F_N^\perp} \rVert_{\text{HS}}^2 = \lVert P_{E_N} - P_{F_N} \rVert_{\text{HS}}^2 < 1. $$ But this implies that $\dim F_N^\perp = \dim E_N^\perp = N$, and therefore $F_N^\perp = \mbox{span}(f_1,\ldots,f_n)$. Which was what we wanted!


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