^{-\frac{1}{2}}u_j\rangle_\mathcal{H} |^2 
  = \sum_{j\in \mathbb{N}}|\langle
  M_\lambda^{-\frac{1}{2}}f,u_j\rangle_\mathcal{H} |^2  \\
& = \langle M_\lambda^{-\frac{1}{2}} S_1 M_\lambda^{-\frac{1}{2}}f,f\rangle_\mathcal{H}
 = \langle S_1 M_\lambda^{-1}  f,f\rangle_\mathcal{H}.
  \end{aligned}
 \end{equation}
Let $f(x) = x \big( \lambda + (1-\lambda )x\big) ^{-1} $ on $[0,\infty
)$. Then $f$ is increasing and $S_1 M_\lambda ^{-1} = f(S_1)$.
Consider the restriction  $S_1:\mathcal{U}\rightarrow \mathcal{U}$ to the subspace $\mathcal{U}$.
 If $\sigma
(S_1) \subseteq [A,B]\subseteq (0,\infty )$, then by the spectral
theorem $\sigma (S_1 M_\lambda ^{-1} ) = \sigma (f(S_1)) \subseteq [f(A),
f(B)]$. 
 Combining this with \eqref{zu3} implies the frame inequality \eqref{zuzeigen}.


Identity  \eqref{zuzeigen1} is proven in Lemma \ref{tight1}.
\end{proof}
\end{lem}

\begin{thm} \label{mix}
Let $\mathcal{T}$ and $\mathcal{U}$ be closed subspaces of a separable Hilbert
space $\H$ such that  $\cos(\varphi_{\mathcal{T},\mathcal{U}})>0$. Let $(u_j)_{j\in \mathbb{N}}$ be a frame for
$\mathcal{U}$ and  $(t_k)_{k\in
  \mathbb{N}}$ be a frame for $\mathcal{T}$. 
For $0< \lambda \leq 1$ let  $L_\lambda$ be  the synthesis operator of the frame
$(u_{\lambda,j} = M_\lambda^{-\frac{1}{2}}u_j)_{j\in\mathbb{N}}$ for
$\mathcal{U}$  and $S_\lambda = L_\lambda L_\lambda^*$ the
corresponding frame operator. 

 Consider the following operators:
\begin{enumerate}[label=(\roman*)]
\item $C_1:=  T (\Sigma_\lambda^{-\frac{1}{2}} U^* T)^\dagger \Sigma_\lambda^{-\frac{1}{2}}$. \label{li}
\item Let the operator $C_2$ be defined on $\R ( U ^* )$  by
\begin{equation}\label{glei}
 C_2  U^* = P_{\mathcal{T},S_\lambda(\mathcal{T})^\perp}
\end{equation}
 and
\begin{equation} \label{gleis}
 C_2 f= 0 \quad \mbox{for }f \in \R( U^*)^\perp.
\end{equation}
Consequently, $C_2$ is independent of the particular choice of the reconstruction frame $(t_k)_{k\in \mathbb{N}}$ for $\mathcal{T}$.
\item  \label{fadi}
For $d\in \ell ^2(\mathbb{N})$ set $C_3 d  = \sum_{k = 1}^\infty
\hat{c}_k t_k$ with
$\hat{c} =
(\hat{c}_{k})_{k\in\mathbb{N}}$ being the minimal norm element of the
set 
\begin{equation*}
K:=\underset{c\in \ell
^2(\mathbb{N})}{\textnormal{arg~min}}~\| U^* T c -d\|_\lambda:=\underset{c\in \ell
^2(\mathbb{N})}{\textnormal{arg~min}}~
\| \Sigma_\lambda^{-\frac{1}{2}} U^* T c - \Sigma_\lambda^{-\frac{1}{2}}
d\|_2^2.
\end{equation*}
\end{enumerate}


Then these operators are equal,  $Q_\lambda := C_1 = C_2 = C_3$,  and $\tilde f = Q_\lambda d$ is the
unique  solution of the least squares problem 
\begin{equation}\label{eq_lambda}
\tilde{f} =  \underset{g\in \mathcal{T}}{\textnormal{arg~min}}
~\| U^* g - d\|_\lambda ^2.  
\end{equation}
\begin{proof}
We apply Theorem~\ref{gensampl} to the frames $( u_{\lambda,j})_{j\in
    \mathbb{N} } $ for $\mathcal{U} $ and $(t_k)_{k\in \N}$ for $\mathcal{T} $. 

Let $L_\lambda ^*$ be the analysis operator of $( u_{\lambda,j})_{j\in
    \mathbb{N} }$ and set $A_1 = T(L_\lambda ^*T)^\dagger$. Since $A_1$ has the equivalent representation (ii) of Theorem~\ref{gensampl} 
$$
A_1 L_\lambda ^* = P_{\mathcal{T}, S_\lambda (\mathcal{T} )^\perp } \quad \text{ on }
\R (L_\lambda ^*)
$$
and $A_1 = 0$ on $\R (L_\lambda ^*)^\perp $. This means that
$$
 P_{\mathcal{T}, S_\lambda (\mathcal{T} )^\perp } = A_1 L_\lambda ^* = T(L_\lambda ^* T)^\dagger L_\lambda ^* =
T(\Sigma_\lambda ^{-\frac{1}{2}} U^* T)^\dagger \Sigma_\lambda ^{-\frac{1}{2}} U^* =
C_1 U^*.
$$
Since $M _\lambda $ is invertible and $L_\lambda ^* = U^* M_\lambda ^{-1/2}$ we have $\R (L_\lambda ^*) =
\R ( U^*M _\lambda ^{-1/2}) = \R (U^*)$ and \eqref{glei} holds for $C_1$. 
To prove that $C_1 = 0$ on $\R (U^*)^\perp $, we use three
algebraic properties of kernels: If $A,B,C $ are bounded, $A$
pseudo-invertible and $C$ invertible, then $\N(A^\dagger) = \N(A^*)$,
$\N(AB) \supseteq \N (B)$ and $\N (AC) = C^{-1} \N (A)$. Consequently
the kernel of $C_1 $ is 
\begin{align*}
  \N (C_1 ) &= \N (T(\Sigma _\lambda ^{-\frac{1}{2}} U^* T)^\dagger \Sigma
  _\lambda ^{-\frac{1}{2}}) \\
&= \Sigma _\lambda ^{\frac{1}{2}} \N (T (\Sigma _\lambda ^{-\frac{1}{2}}  U^* T)^\dagger) 
\supseteq \Sigma _\lambda ^{\frac{1}{2}}\N  ((\Sigma _\lambda ^{-\frac{1}{2}} U^* T)^\dagger) \\
&= \Sigma _\lambda ^{\frac{1}{2}} \N  (T^*U\Sigma _\lambda ^{-\frac{1}{2}}) =
  \N (T^*U \Sigma_\lambda ^{-1}) \\
& \supseteq  \N (U \Sigma _\lambda ^{-1} ) =  \R (\Sigma _\lambda ^{-1}
U^*)^\perp \\
& =\R (U^* M_\lambda ^{-1})^\perp  = \R (U^*)^\perp \,.
\end{align*}
Thus $C_1 = 0$ on $\R (U^*)^\perp $, which is \eqref{gleis}. 

For showing $C_1 = C_3$ and \eqref{eq_lambda} we repeat the proof of
Theorem~\ref{wichtig} verbatim.~\end{proof} 

\end{thm}


The following Lemma gives a useful upper bound on the quasi-optimality constant of the operators $Q_\lambda$.

\begin{lem} \label{quasi_constant}
Let $\cos(\varphi_{\mathcal{T},\mathcal{U}})>0$ and let $Q_\lambda$ be defined as in Theorem \ref{mix}.
Then the quasi-optimality constant $\mu(Q_\lambda) = \|Q_\lambda U^*\|_{\rm op}$ is bounded by 
\begin{equation}\label{eq47}
\|Q_\lambda U^*\|_{\rm op}\leq \frac{1}{\cos(\varphi_{\mathcal{T},\mathcal{U}})}\sqrt{\frac{B(\lambda +A(1-\lambda))}{A(\lambda + B(1