Boundedness and Operator Norms
[7/100%]Prove that is a normed linear space. That is, prove Theorem 5.3.
Theorem 5.3: Let and be normed linear spaces. The collection of all bounded linear operators , endowed with the operator norm, forms a normed linear space.
Show that the identity operator from to is a bounded linear operator, but that the identity map from to is unbounded. This phenomenon cannot happen on a Banach space; see Exercise 6.3.2.
Fix a continuous function and define the composition operator by . Prove that is a bounded linear operator with .
For define
Show that is a bounded linear operator from to , and compute .
Consider an matrix that satisfies for all pairs of indices and . Such a matrix is called Hermitian. View as a linear operator . Assume that is endowed with the usual norm. Prove that , where are the eigenvalues of .
Consider
as a linear operator from to itself. The aim of this exercise is to show that the operator norm of depends on the choice of norm on .
Endow with the supremum norm, . Show that in this case, the norm of is given by
Endow with the 1-norm, . Show that in this case, the norm of is given by
Endow with the 2-norm, . Consider the matrix
and form the matrix . Show that in this case, the norm of is given by
Use this to write down a formula for the norm of as a function of its entries . Here "tr" denotes trace, and "det" denotes determinant.
Finally, compute these 3 operator norms of
Let denote the collection of all continuous functions with continuous derivative (including one-sided derivatives at the endpoints).
Show that , with the supremum norm, is a subspace of but that it is not closed.
Prove that the differential operator
is linear but not bounded.