A good question. It seems to me that the

most natural approach is to consider G=T*T.

Then ||G|| = ||T||2, G is self-adjoint and also

does not attain its norm. After that one can use

the general theory of self-adjoint operators

(similarity with an operator of multiplication

by a function). This will lead to the solution.

Yes, you can build the sequence as follows, 

Pick the first vector e_1 such that ||Te_1|| > ||T|| - 1/2

Let E_1 = span{e_1} and decompose H = E_1 + H_1 where H_1 is orthogonal to E_1

Now look at T on H_1.  It is still linear bounded operator and the norm is still ||T|| and it is not achieved,

so pick the second vector e_2 such that ||Te_2|| > ||T|| - 1/2^2 and continue the procedure,

The important fact is: if H = M1 + M2 then ||T|| = max (||T on M1|| , ||T on M2|| )

Dear Professor Taghvaei,

Thanks for your answer. But how to prove that 

if H = M1 + M2 then ||T|| = max (||T on M1|| , ||T on M2|| )

for a Hilbert space.

The property "if H = M1 + M2 then ||T|| = max (||T on M1|| , ||T on M2|| )" is not correct.

As a counterexample consider the operator T on R^2 (which is also a Hilbert space) that maps every vector (a, b) into vector (a+b, 0). Let M1 be the subspace of vectors of the form (a, 0) and M1 be the subspace of vectors of the form (0, b). Then

||T|| = \sqrt{2}, but  max (||T on M1|| , ||T on M2|| ) = 1.

Right. So the method does not work.

Suppose that T is  a compact operator. Then for any orthonormal sequence {e_n}  ||Te_n|| tends to 0.

Thanks @Alexei. 

So in general there does not exist a sequence \{e_n\} of orthonormal vectors such that 

\|Te_n\| --> \|T\|. At least the operator has to be non-compact.

@ Alexei, Since the operator in question does not attain norm, so T is non-compact. 

Thus the question still remains to be answered

Suppose T is a bounded (non-compact)   linear operator on a Hilbert space H such that T does not attain its norm i.e., there does not exist x in H with ||x||=1 such that ||Tx||=||T||. Then can we say that there exists a sequence {e_n} of orthonormal vectors such that ||Te_n|| tends to ||T||?

Dear Professor Kallol Paul,

Maybe I did not express my idea clearly,

but the question IS answered. In my first

post I gave a hint how to reduce the problem

to self-adjoint operators, and by general

similarity theorem (Spectral theorem in terms

of multiplication operator, see Michael Reed,

Barry Simon. Methods of modern mathematical physics,

v. 1: : Functional analysis, Corollary after Theorem VII.3)

a self-adjoint operator is unitarily equivalent to

an operator of multiplication by a function F on

an appropriate $L_2(\mu)$. For the operator of

multiplication by F the answer is positive (and easy),

so it is also positive in general.

All the best, Vladimir Kadets

Thanks a lot Professor Kadets.

Clearly, there exists a sequence {e_n} of orthonormal vectors such that ||Te_n|| tends to ||T||  if and only if ||T|| lies in the essential spectrum of the operator T.  Therefore the answer is really easy but not positive in general. 

I'm sorry.  there's a misprint in my previous message: 

It is true for a self-adjoint operator T.

My bad, I was wrong, the fact that "if H = M1 + M2 then ||T|| = max (||T on M1|| , ||T on M2|| )" is not true in general, It is true for self adjoint operators,

To complete the last step of Professor Kadets answer, we obtain orthonormal x_n s.t ||TT^*x_n|| -> ||T||^2, and then choose e_n= T^*x_n/||T^*x_n||

I should add that compact operators attain their norms. In particular Professor Kadets give a complete (positive) answer on the question.

Dear Professor Taghvaei,

unfortunately, your statement

"if H is the orthogonal sum of M1 and M2 then

for self-adjoint operator T on H

||T|| = max (||T on M1|| , ||T on M2|| )"

is also not correct. As a counterexample consider

again H = R^2, M1being the subspace of vectors

of the form (a, 0) and M2 - the subspace of vectors

of the form (0, b).

Let T be the operator which maps (a,b) to (a+b, a+b).

Then T is self-adjoint, ||T|| = 2, but the restrictions

of T onto M1 and M2 both have norm \sqrt{2}.

To make the formula correct one needs the assumption that

T(M1) and T(M2) are mutually ortogonal. This happens, for example,

when M1 and M2 are invariant subspaces of T.

Thanks a lot @ Alexei and @ Kadets. 

The answer to my question is yes.  We donot  use the Spectral theorem. The details are as follows:

We note that $T^{*}T$ is a positive operator, $ \|T^{*}T \| = \|T\|^2 $ and $T$ attains norm if and only if $T^{*}T$ attains norm. Also $ \|T\|^2 \in \sigma(T^{*}T) = \sigma_{disc}(T^{*}T) \cup \sigma_{ess}(T^{*}T).$ By Theorem VII.10 of Reed and Simon ( Methods of modern mathematical physics vol.I) , $\|T\|^2$ can't belong to $ \sigma_{disc}(T^{*}T) $. So by Theorem VII.12(Weyl's criterion) of the same book, there exists $\{e_n\} $ of orthonormal vectors such that $ \| (T^{*}T - \|T\|^2)e_n \| --> 0.$

Now $ \mid \langle (T^{*}T - \|T\|^2)e_n, e_n \rangle \mid \leq \| (T^{*}T - \|T\|^2)e_n \| --> 0.$

This implies $ \langle (T^{*}T - \|T\|^2)e_n, e_n \rangle = \langle T^{*}Te_n, e_n \rangle - \langle \|T\|^2 e_n, e_n\rangle --> 0.$ Thus $ \|Te_n\| -->\|T\|.$