Consider a linear Hermitian operator acting on a finite dimensional vector space. This operator is bounded, and in order to find the bound I am calculating the largest eigenvalue in its matrix representation.
Now if my vector space is an infinite dimensional and continuous Hilbert space (a non-separable Hilbert space, say $l^2(\mathbb{R})$), do I have any similar notion of "largest eigenvalue" of the operator? Or is there another way to determine the bound of this operator?
Note: I have a physics background, would appreciate some examples.
If you have a selfadjoint bounded operator $T$ on a Hilbert space, you always have $$ \|T\|=\max\{|\lambda|:\ \lambda\in \sigma(T)\}, $$ where $\sigma(T)$ is the spectrum $$ \sigma(T)=\{\lambda:\ T-\lambda I\ \text{ is not invertible}\}. $$ The equality holds more generally when $T$ is normal ($T^*T=TT^*$) but fails in general: for example $T=\begin{bmatrix} 0&1\\0&0\end{bmatrix}$ has $\|T\|=1$ but $\sigma(T)=\{0\}$.
You cannot do the above with eigenvalues, as these may fail to exist. For example let $H=L^2[0,1]$, and $T$ the multiplication operator $$ (Tf)(x)=xf(x). $$ Then $\|T\|=1$ and $\sigma(T)=[0,1]$, but $T$ has no eigenvalues.