# Answer on Functional Analysis Question for sharmistha saha

Question #4653

1)Give one example of a normal operator on a hilbert space which is not unitary.Justify.

2)Give one example of a normed space which is not an inner product.Justify.

2)Give one example of a normed space which is not an inner product.Justify.

Expert's answer

1) By definition a normal operator on a complex Hilbert space H is a continuous

linear operator

N:H --> H

that commutes with its hermitian adjoint

N*:

N N* = N* N.

A unitary operator is an operator U:H --> H

satisfying the identity

& U U* = U* U = I.

Therefore if U is any

unitary operator, e.g the identity U=I, and

a =/= 1 is an arbitrary complex

number distinct from 1, then&

& N=aU

is normal.

Indeed,& N* = aU*,

whence

N N* = (a U) (a U*) = a^2 (U U*) = a^2 (U* U) = (a U*) (a U) = N*

N.

2) Fix p>=1 and consider the space l^p consisting of all infinite

sequences

x = (x1, x2, ... )

such that

sum_{i=1}^{infinity}

|xi|^p < infinity

Then l^p is a hilbert space only for p=2.

For all

other p it is a normes space but not an inner space.

Actually, for a

norm |*| on a linear space V to be induced by inner product <*,*>, so

|x|^2 = <x,x>

it is necessary and sufficient that

|x+y|^2 +

|x-y|^2 = 2(|x|^2 + |y|^2)

for any x,y from V.

linear operator

N:H --> H

that commutes with its hermitian adjoint

N*:

N N* = N* N.

A unitary operator is an operator U:H --> H

satisfying the identity

& U U* = U* U = I.

Therefore if U is any

unitary operator, e.g the identity U=I, and

a =/= 1 is an arbitrary complex

number distinct from 1, then&

& N=aU

is normal.

Indeed,& N* = aU*,

whence

N N* = (a U) (a U*) = a^2 (U U*) = a^2 (U* U) = (a U*) (a U) = N*

N.

2) Fix p>=1 and consider the space l^p consisting of all infinite

sequences

x = (x1, x2, ... )

such that

sum_{i=1}^{infinity}

|xi|^p < infinity

Then l^p is a hilbert space only for p=2.

For all

other p it is a normes space but not an inner space.

Actually, for a

norm |*| on a linear space V to be induced by inner product <*,*>, so

|x|^2 = <x,x>

it is necessary and sufficient that

|x+y|^2 +

|x-y|^2 = 2(|x|^2 + |y|^2)

for any x,y from V.

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