Question on the self-adjointness of an operator

Solution 1:

The domain of $T^*$ is $$ D(T^*)=\{y\in H:\ x\longmapsto\langle Tx,y\rangle\ \text{ is bounded on }D(T) \}. $$ So we want to analyze the boundedness of the map $$ \phi(x)=\langle Tx,y\rangle=\sum_k\lambda_kx_ky_k,\qquad x\in D(T). $$ The sum is finite because $x$ has finitely many nonzero terms. Using Cauchy Schwarz, $$ |\phi(x)|^2\leq\|x\|^2\,\sum_k|\lambda_k|^2\,|y_k|^2. $$ So $$ D(T^*)\supset\{y:\ \sum_k|\lambda_k|^2\,|y_k|^2<\infty\}. $$ Conversely, if $y\in D(T^*)$ then there exists $c>0$ such that $|\langle Tx,y\rangle|\leq c$ for all $x\in D(T)$ with $\|x\|=1$. Being bounded, the functional extends to all of $H$. Then $$ \sum_k|\lambda_k|^2\,|y_k|^2=\sup\Big\{\Big|\sum_kx_k\,\lambda_ky_k\Big|:\ \sum_k|x_k|^2=1\Big\}=\sup\{|\langle Tx,y\rangle|:\ \|x\|=1\}\leq c. $$ Thus $$ D(T^*)=\{y:\ \sum_k|\lambda_k|^2\,|y_k|^2<\infty\} $$ and $$ T^*y=\sum_k\lambda_ky_ke_k. $$

Now suppose that $x\in D(T^*)$ and $(T^*+i)x=0$. That is, $T^*x=-ix$. The fact that $x\in D(T^*)$ guarantees that $\sum_k\lambda_kx_ke_k\in H$. The equality $T^*x=-ix$ is $$ \sum_k\lambda_kx_ke_k=-i\sum_kx_ke_k. $$ Comparing coordinate-wise, this gives us $\lambda_kx_k=-ix_k$. Because $\lambda_k$ is real, this equality can only happen when $x_k=0$. Being the case for all $k$, we get that $x=0$ and so $T^*+i$ is injective.