Given a linear operator between Hilbert spaces $T:X \to Y$ where $X$ is infinite dimensional I wonder if there is a method to compute its norm using an orthonormal basis $\{e_i:i \in I\}$. I inspired myself in this answer in where they seem to be using such a method.

Could I apply that method to$$T(f) = \pi \int_0^{2\pi} f(x)\,dx + \frac{i}{2} \int_\pi^{2\pi} \frac{f(x)}{x} \,dx$$ knowing that $L^2[-\pi,\pi]$ has the following orthonormal basis?$$\left\{\frac{1}{\sqrt{2}^n}, \frac{\cos t}{\sqrt{n}}, \frac{\sin t}{\sqrt{n}}, \frac{\cos 2t}{\sqrt{n}}, \cdots \right\}$$

  • $\begingroup$ General normed spaces $X,Y$ are irrelevant here, as there is a notion of orthogonality only in inner-product spaces, so that the natural framework for such questions are operators defined on Hilbert spaces. $\endgroup$ – uniquesolution Feb 8 '18 at 12:01
  • $\begingroup$ @uniquesolution thank you i edit my question $\endgroup$ – Rodrigo Feb 8 '18 at 12:03

There is no such method. Even in finite dimension, knowing the canonical basis doesn't help you in finding the norm of a $10\times 10$ matrix $M$, where the "easy" method is to take the square root of the biggest root of the degree-$10$ polynomial $\det(M^*M-\lambda I)$.

When you can find the norm of an operator, it is almost surely due to some ad-hoc estimate, or the operator being very easy (i.e., normal).


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