Sequence of operators in $L^2(\mathbb{R})$ I am trying to solve the following exercise:
In the space $L^2(\mathbb{R})$ define a sequence of operators $(A_n)$ that are such that $||A_nf||=||f||$ and $\langle A_nf,g\rangle\to0$ for every $f,g\in L_2(\mathbb{R})$.
I have thought of defining $(A_n)f(x)=f(x+n)$ but I don't know how to proceed, can someone give me a hint? Thanks a lot.
 A: It seems like this question has had many iterations, but I'll react to what I see now in the hopes that I am right. My answer would be to essentially look at $A_n$, where it is  the $n$-shift operator on $\ell^2_{\mathbb{R}}(\mathbb{N})$.
Take a countable orthonormal basis of $L^2(\mathbb{R})$, and denote it by $\{ \phi_n\}_{n=1}^\infty \subseteq L^2(\mathbb{R}) $. For every $f\in L^2(\mathbb{R})$ there is a unique sequence $\{ a_n(f) \}_{n=1}^\infty$ such that
$$ f=\sum_{k=1}^\infty a_k(f)\cdot \phi_k.$$
$A_n(f)$ will simply be
$$ A_n(f)=\sum_{k=1}^\infty a_{k}(f)\cdot \phi_{n+k}. $$
Notice that
$$ \langle A_n(f),\phi_{k} \rangle= \begin{cases} a_{k-n}(f) & ,k> n\\ 0 & ,k\leq n \end{cases}  $$
Hence, $\langle  A_n(f), \phi_k\rangle=0$ for $n$ large enough.
By Paresval's identity, we know that
$$\Vert A_n(f) \Vert^2= \sum_{k=1}^\infty \vert a_k(f)\vert^2. $$
Any finite linear combination
$$ g' = \sum_{j=1}^N c_j \cdot \phi_{k_j}, $$
satisfies that
$$  \langle A_n(f) ,\sum_{j=1}^N c_j \cdot \phi_{k_j}\rangle=0  $$
for $n$ large enough. Since the closure of the linear span of $\{ \phi_n \}$ is equal to $L^2$, you can deduce that
$$ \langle A_nf,g\rangle\to0 $$
for all $g\in L^2(\mathbb{R})$. Notice that $f$ was arbitrarily chosen here.
A: I thaught of taking
$$
A_n f=
\begin{cases}
f(x-n),\quad \text{if $x\ge n$}\\
0,\qquad \qquad\text{if $-n<x<n$}\\
f(x+n), \quad \text{if $x\le -n$}
\end{cases}
$$
It is an isometry, so $||A_n f||=||f||$.
$\langle A_n f,g\rangle=\int_{-\infty}^{-n} f(x+n)\overline{g(x)}dx + \int_{n}^{+\infty} f(x-n) \overline{g(x)}dx$.
The two integrals are tails of convergent integrals (since $\langle A_n f,g\rangle<\infty$), so they tend to zero as $n\to +\infty$
