Characterization of Parseval Frame

Question

Let $H$ be a Hilbert space, and let $e_j\in H$ for every $j\in\mathbb N$.

Is it possible to show that

$$\left\|f\right\|^2=\sum_{j=1}^\infty\left|\langle f,e_j\rangle\right|^2\text{ for every }f\in H\text{ if and only if }f=\sum_{j=1}^\infty\langle f,e_j\rangle e_j\text{ for every }f\in H\tag*{?}$$

I get the feeling that we need $e_j\perp e_k$ if $j\neq k$.

Answer 1

For the reverse implication see this answer.

Note that orthogonality is not required for a Parseval Frame.

Assume that $\|f\|^2=\sum_j|\langle f,e_j\rangle|^2$ for all $f$. Given $m>n$, \begin{align} \Big\|\sum_{j=1}^m\langle f,e_j\rangle\,e_j-\sum_{j=1}^n\langle f,e_j\rangle e_j\Big\|^2 &=\Big\|\sum_{j=n+1}^m\langle f,e_j\rangle e_j\Big\|^2\\[0.3cm] &=\sup\Big\{\Big|\Big\langle\sum_{j=n+1}^m\langle f,e_j\rangle e_j,h\Big\rangle\Big|:\ \|h\|=1\Big\}\\[0.3cm] &=\sup\Big\{\Big|\sum_{j=n+1}^m\Big\langle\langle f,e_j\rangle e_j,h\Big\rangle\Big|:\ \|h\|=1\Big\}\\[0.3cm] &\leq\sup\Big\{\sum_{j=n+1}^m|\langle f,e_j\rangle\,\langle e_j,h\rangle|:\ \|h\|=1\Big\}\\[0.3cm] &\leq\bigg(\sum_{j=n+1}^m|\langle f,e_j\rangle|^2\bigg)^{1/2} \sup\Big\{\bigg(\sum_{j=n+1}^m|\langle e_j,h\rangle|^2\bigg)^{1/2}:\ \|h\|=1\Big\}\\[0.3cm] &\leq\bigg(\sum_{j=n+1}^m|\langle f,e_j\rangle|^2\bigg)^{1/2} \sup\Big\{\bigg(\sum_{j=1}^\infty|\langle e_j,h\rangle|^2\bigg)^{1/2}:\ \|h\|=1\Big\}\\[0.3cm] &=\bigg(\sum_{j=n+1}^m|\langle f,e_j\rangle|^2\bigg)^{1/2}. \end{align} The convergence of $\sum_j|\langle f,e_j\rangle|^2$ guarantees that the last series can be made arbitrarily small if $m$ and $n$ are large enough. So $$ \sum_j\langle f,e_j\rangle e_j $$ exists. Now we need to show that it equals $f$. We will need the Polarization Identity $$ \langle f,g\rangle=\tfrac14\,\sum_{k=0}^3i^k\|f+i^kg\|^2, $$ and also the particular case to complex numbers, $$ z\overline w=\tfrac14\,\sum_{k=0}^3i^k|z+i^kw|^2. $$ We have \begin{align} \langle f,g\rangle&=\tfrac14\,\sum_{k=0}^3i^k\|f+i^kg\|^2 =\tfrac14\,\sum_{k=0}^3i^k\sum_j|\langle f+i^kg,e_j\rangle|^2\\[0.3cm] &=\sum_j\tfrac14\,\sum_{k=0}^3i^k|\langle f,e_j\rangle+i^k\langle g,e_j\rangle|^2\\[0.3cm] &=\sum_j\langle f,e_j\rangle\langle e_j,g\rangle. \end{align} Then \begin{align} \Big\langle f-\sum_j\langle f,e_j\rangle e_j,g\Big\rangle &=\langle f,g\rangle-\sum_j\langle f,e_j\rangle\langle e_j,g\rangle=0. \end{align} As this can be done for any $g$, it follows that $$ f=\sum_j\langle f,e_j\rangle e_j. $$

Answer 2

If $(e_j)$ is a sequence in $H$ with $<e_j,e_k>=\delta_{jk}$, then we have:

$(e_j)$ is an orthonormalbasis of $H$ $ \iff$

$\left\|f\right\|^2=\sum_{j=1}^\infty\left|\langle f,e_j\rangle\right|^2\text{ for every }f\in H \iff\sum_{j=1}^\infty\langle f,e_j\rangle e_j\text{ for every }f\in H\tag*{}$

Answer 3

No, you don't need to assume orthogonality. The claim becomes more clear in terms of the analysis operator (aka Bessel operator) $$ Af = (\langle f, e_j\rangle)_j $$ and the synthesis operator (aka reconstruction operator) $$ S(c_j) = \sum_j c_j e_j $$

The claim is that $SA=I$ if and only if $A$ is an isometry into $\ell^2$.

Proof or $\impliedby$. Suppose $A$ is an isometry into $\ell^2$. Then its adjoint $A^*$ is a bounded operator from $\ell^2$ to $H$, which agrees with $S$ because $$\langle Af, x\rangle_{\ell^2} =\sum_j \langle f, e_j\rangle_H \overline{x_j} = \left\langle f, \sum_j x_je_j\right\rangle_H = \langle f, Sx\rangle_H $$ By the polarization identity, $\langle Af, Ag\rangle_{\ell^2} = \langle f, g\rangle_H $ for all $f,g\in H$. Hence $\langle A^*Af, g\rangle_H = \langle f, g\rangle_H $ for all $f,g\in H$, which implies $A^*A=I$. Since $A^*=S$, we conclude that $SA=I$.

Proof of $\implies$. Here I have to assume that $A$ is a bounded operator into $\ell^2$; it may be possible to remove this assumption, but I don't see how; in any case it is standard in frame theory. As above, we have $S=A^*$. Since $SA=I$, it follows that $A*Af=f$ for every $f\in H$, hence $$ \|Af\|^2 = \langle Af, Af\rangle_{\ell^2} = \langle A^*Af, f\rangle_H = \langle f, f\rangle _H = \|f\|^2 $$ as desired.