An example of the Pseudo-inverse of an operator Let $E$ an infinite dimensional complex Hilbert space and $\mathcal{L}(E)$  be the algebra of all bounded linear operators on $E$.
Definition: Let $T \in \mathcal{L}(E)$. The Moore-Penrose inverse of $T$, denoted by $T^{+}$, is defined as the unique linear extension of $(\overline{T})^{-1}$ in
 $$D(T^{+}) = \mathcal{R}(T)+\mathcal{R}(T)^{\perp},$$
with $\mathcal{N}(T^{+}) = \mathcal{R}(T)^{\perp}$ and  $\overline{T}$ is the isomorphism
$$\overline{T}:=T|_{{\mathcal{N}(T)}^{\perp}}: {\mathcal{N}(T)}^{\perp} \longrightarrow \mathcal{R}(T).$$
Moreover, $T^{+}$ is the unique solution of the four ''Moore-Penrose equations'':
$$TXT = T,\quad XTX = X,\quad XT = P_{N{(T)^{\bot}}}\,\,\mbox{and}\,\,\quad TX = P_{\overline{\mathcal{R}(T)}}{{|}_{D(T^{+})}}.$$
Here $\mathcal{R}(T)$ and $\mathcal{N}(T)$ denote respectively the range and the nullspace of $T$. Also $P_{F}$ denote the orthogonal projection onto $F$.

I want to see with an example how we compute $T^{+}$ when $T$ is a non invertible operator acting on an infinite dimensional complex Hilbert space $E$.

 A: If you take $T\in B(\ell^2(\mathbb N))$ given by $Te_n=\tfrac1n e_n$ on the canonical orthonormal basis, that is 
$$
T(x_1,x_2,\ldots)=(x_1,\tfrac12 x_2,\tfrac13 x_3,\ldots), 
$$
then 
$$
T^+(x_1,x_2,\ldots)=(x_1,2 x_2,3 x_3,\ldots).
$$
A very natural example is given by the shift: if $S$ is the unilateral shift
$$
S(x_1,x_2,\ldots)=(0,x_1, x_2, x_3,\ldots), 
$$
then one can check directly from the Moore-Penrose equations that $S^+=S^*$, i.e., 
$$
S^+(x_1,x_2,\ldots)=(x_2,x_3,\ldots). 
$$
A: The following example is a solution to an exercise from here (in German, taken from lectures at LMU Munich, 2007).

For $a < b \in \mathbb{R}$ let
      \begin{equation*}
    T: L^2(a,b) \to \mathbb{R}, \ 
    f \mapsto \int_{a}^{b} f(x) dx.
\end{equation*}

We have
    \begin{equation*}
    \ker(T)
    = \left\{ f \in L^2: \int_{a}^{b} f(x) \cdot \mathbb{1}_{(a,b)}(x)  dx = 0 \right\}
    = \{ \mathbb{1}_{(a,b)} \}^{\perp}
\end{equation*}
    thus
    \begin{equation*}
    \ker(T)^{\perp} 
    = \{ \mathbb 1_{(a,b)} \}^{\perp \perp}
    = \{ f: (a,b) \to \mathbb K, f(x) = c \in \mathbb{R} \ \forall x \in (a,b) \}, 
\end{equation*}
    i.e. $\ker(T)^{\perp}$ consists of all constant functions on $(a,b)$.
    We have $\text{ran}(T) = \mathbb{R}$ and thus $\text{ran}(T)^{\perp} = \{0\}$.
    Therefore
    \begin{equation*}
    \widetilde{T}: \ker(T)^{\perp} \to \mathbb{R}, \ 
    \big(f(x) = c\big) \mapsto \int_{a}^{b} f(x) dx
    = (b - a) c,
\end{equation*}
    implying
    \begin{equation*}
    T^+ = \widetilde{T}^{-1}: \mathbb{R} \to \ker(T)^{\perp}, \ 
    c \mapsto \left(f(x) := \frac{c}{b-a}\right).
\end{equation*}

Now let $w = (w_i)_{i = 1}^{n} \subset L^2(a,b)$ be an orthonormal system and consider
      \begin{equation*}
    T: L^2(a,b) \to \mathbb{R}^n, \ 
    f \mapsto \left( \int_{a}^{b} f(x) w_i(x) dx\right)_{i = 1}^{n}.
\end{equation*}

As above, $\text{ran}(T) = \mathbb{R}^n$.
    Furthermore
    \begin{equation*}
    \ker(T)
    = \{ f \in L^2(a,b): \langle f, w_i \rangle = 0 \ \forall i \in \{1, \ldots, n\}\} 
    = w^{\perp}
\end{equation*}
    and thus $\ker(T)^{\perp} = w^{\perp \perp} = \text{span}(w)$.
    Then
    \begin{align*}
    \widetilde{T}: \text{span}(w) \to \mathbb{R}^n, \ 
    \sum_{k = 1}^{n} a_k w_k(x)
    & \mapsto \left( \int_{a}^{b} \sum_{k = 1}^{n} a_k w_k(x) w_i(x) dx \right)_{i = 1}^{n} \\
    & = \left( \sum_{k = 1}^{n} a_k \langle  w_k, w_i \rangle \right)_{i = 1}^{n} 
    = (a_i)_{i = 1}^{n},
\end{align*}
    holds as $\langle w_i, w_j \rangle = \delta_{i,j}$ since $w$ is orthonormal.
    Thus
    \begin{equation*}
    T^+ = \widetilde{T}^{-1}: \mathbb{R}^n \to \text{span}(w), \ 
    (a_i)_{i = 1}^{n} \mapsto \sum_{k = 1}^{n} a_k w_k(x)
\end{equation*}
