In finite dimensions, the numerical range $$W(T) := \left\{ \frac{x^* T x}{x^* x}: x \in \mathbb{C}^n \setminus \{0\}\right\}$$ of an matrix $T \in \mathbb{C}^{n \times n}$ contains all eigenvalues of $T$.
In infinite dimensions (a complex Hilbert space $(\mathcal{H}, \langle \cdot, \cdot \rangle$) we define the numerical range of a linear continuous operator $T \in L(\mathcal{H})$: $$ W(T) := \{ \langle Tx, x \rangle: \| x \| = 1\}. $$ Let $\sigma(T)$ denote the spectrum of $T$, then we know that $\sigma(T) \subset \overline{W(T)}$ (we actually also have $\sigma(T) \subset W(T)$ in finite dimensions). Does $W(T)$ also contain all eigenvalues in this case or does only $\overline{W(T)}$ contain all eigenvalues?
My ideas: Let $\lambda \in \mathcal{C}$ be an eigenvalue. Then there exists a vector $x \ne 0$ such that $Tx = \lambda x$. Now $\tilde{x} := \frac{1}{\| x \|} x$ has unit norm and we have $$ \langle T \tilde{x}, \tilde{x} \rangle = \frac{1}{\| x \|^2} \langle Tx, x \rangle = | \lambda |. $$
Is my proof correct or do I have to get rid of the absolute value somehow?
Your computation is wrong. $\langle Tx,x \rangle=\lambda \langle x,x \rangle$ so you actually get $\lambda$ instead of $|\lambda|$.