Construct space and measure on it in order to make unitary equivalence

Question

Let $A : l_2 \to l_2 \quad A(x_1,\dots ,x_n ,\dots) = (\lambda_1 x_1,\dots,\lambda x_n,\dots)$ where all $\lambda_i$ different and $\{\lambda_j\}_{j \in \mathbb{N}}$ - bounded sequence of real numbers. I need to pick up segment $[a,b]$ and measure on it in order to make operator $A$ unitary equivalent to multiplication operator by function $t$ in space $L_2([a,b] , \mu)$.

Answer 1

Hint: unitarily equivalent operators have the same spectrum. The spectrum of a multiplication operator is the essential range of it's corresponding function. So, it is only reasonable to think as follows:

Since the operator $A:\ell^2\to\ell^2$, $(x_n)\mapsto(\lambda_nx_n)$ has spectrum $\overline{\{\lambda_n:n\in\mathbb{N}\}}$, consider any interval $[a,b]$ such that $\{\lambda_n:n\in\mathbb{N}\}\subset(a,b)$. Then, define the measure $\delta$ on $\mathcal{P}([a,b])$ by $\delta(E)=\text{number of $\lambda_i$ contained in $E$}.$ Then, you can prove that $A$ is unitarily equivalent with $M_t$ where $t$ is the identity function on $[a,b]$ and $M_t:L^2([a,b],\delta)\to L^2([a,b],\delta)$ is the multiplication operator by $t$.

Edit: I am adding the construction of the unitary operator. It is important to keep in mind that in $L^p$ spaces two functions $f,g$ are equal if and only if they are equal almost everywhere. Therefore, in $L^2([a,b],\delta)$, two functions are equal if and only if $f(\lambda_i)=g(\lambda_i)$ for all $i$.

We now define $U:\ell^2\to L^2([a,b],\delta)$ given by $U(x_n)_{n=1}^\infty=f$, where $f(\lambda_i)=x_i$ for all $i$ and $f=0$ everywhere else. The operator is surjective: Indeed, let $f$ be a function of $L^2([a,b],\delta)$. Then we have $\int_{[a,b]}|f(x)|^2d\delta(x)<\infty$, but by the way we defined the measure $\delta$, we see that this integral is simply $\sum_{i=1}^\infty|f(\lambda_i)|^2$. Since this series converges, the sequence $(f(\lambda_i))_{i=1}^\infty$ is an element of $\ell^2$ and it is mapped to $f$ through $U$. The operator preserves inner product: Simply notice that for $f,g\in L^2([a,b],\delta)$ it is $\langle f,g\rangle=\int_{[a,b]}f(x)\overline{g(x)}d\delta(x)=\sum_{i=1}^\infty f(\lambda_i)\overline{g(\lambda_i)}$. Now it is immediate that $U$ preserves inner product.

Since surjective+preserving inner product = unitary, we are done.

I leave it for you to verify that $UAU^{-1}=M_t$.