Hello. Could you please help me to solve this problem:
is a linear unbounded operator on Hilbert space . If AU=UA for all - unitary operator on , then is bounded operator and .
Fix a unit vector . Let be another unit vector, orthogonal to both and . Let x and y be an arbitrary pair of unit vectors orthogonal to each other. Then there exists a unitary U with and . Therefore Similarly,
Thus Ax is orthogonal to everything in the orthogonal complement of x, so that Ax must be a multiple of x. What's more, that multiple must be the same for x as it is for . Conclusion: A is a scalar multiple of the identity operator.
Let V be the two-dimensional subspace spanned by and , and let W be the two-dimensional subspace spanned by and . Then U, as defined above, maps V isometrically onto W. Also, the subspaces and have the same dimension, and so they are isometrically isomorphic to each other. Define U to be a mapping implementing this isomorphism. Then U, defined in that way, is an isometric isomorphism from H onto H, hence a unitary operator.