Proving that an operator is self-adjoint.

Question

Let $(\lambda_{n})_{n \in \mathbb{N^{*}}}$ be a sequence of real numbers converging to $0$ and let $(u_{n})_{n \in \mathbb{N^{*}}}$ be an orthonormal family in a Hilbert space $H$. Define $T:H \rightarrow H$ by $$T(u) = \sum_{k=1}^{\infty} \lambda_{k}\langle u,u_{k}\rangle u_{k}$$

Prove that $T$ is self-adjoint. could anyone help me please in doing so?

Thanks!

Answer 1

Hint:

$$T\text{ is self adjoint }\Leftrightarrow\langle Tx,x\rangle\in\mathbb{R}\text{ for every }x\in H.$$

Answer 2

Just use linearity and continuity of the inner product.

For any $x, y \in H$ we have:

\begin{align} \langle Tx, y\rangle &= \left\langle \sum_{k=1}^\infty \lambda_k \langle x, u_k\rangle u_k, y\right\rangle \\ &= \sum_{k=1}^\infty \lambda_k \langle x, u_k\rangle \left\langle u_k, y\right\rangle \\ &= \left\langle x, \sum_{k=1}^\infty \overline{\lambda_k\langle u_k, y\rangle} u_k\right\rangle \\ &= \left\langle x, \sum_{k=1}^\infty \lambda_k \langle y, u_k\rangle u_k\right\rangle \\ &= \langle x, Ty\rangle \end{align}

Hence $T^* = T$.