In mathematics, an operator in a complex or real Hilbert space is called Hermitian, symmetric, if it satisfies the equality for all in the domain of definition . Here and below it is assumed that — is the inner product in . The name is given in honor of the French mathematician Charles Hermite.

An operator in is called self-adjoint, or hypermaximal Hermitian, if it coincides with its adjoint.

Adjoint operator — a generalization of the concept of the Hermitian-adjoint matrix for infinite-dimensional spaces.A self-adjoint operator is symmetric; the converse, generally speaking, is not true. For continuous operators defined on the whole space, the notions of symmetric and self-adjoint coincide.

Definition 4. A bounded linear operator A in a Hilbert space H is called self-adjoint or symmetric, if it coincides with its adjoint: A = A*.

In other words, a self-adjoint operator A is characterized by the condition (Ax, y) = (x, Ay) for . In the last example, if the kernel K(t, s) is symmetric: K(t, s) = K(s, t), then

and hence, the integral operator will be symmetric.

It is easy to see that any linear combination of self-adjoint operators is also a self-adjoint operator.

Thus, in the normed linear space of linear operators mapping H into H, the self-adjoint operators form a linear manifold. Moreover, we shall now prove that this subset is closed and, consequently, is a subspace. In other words, if the operators An – are self-adjoint and An (in norm), then the operator A – is also self-adjoint. We shall prove an even stronger statement.

Theorem 13. If the operators An – are self-adjoint and the sequence {An} converges pointwise to an operator A, then A will also be a self-adjoint operator.

Proof. It follows from the continuity of the inner product that for any .

(Ax, y) = ( Anx, y) = (Anx, y) = (x, Any) = (x, Any) = (x, Ay).

The theorem is proved.

If the operators A and B – are self-adjoint, then Consequently, for the operator AB to be self-adjoint, it is necessary and sufficient that , i.e., that the operators A and B commute with each other. In particular, all powers of a self-adjoint operator A are also self-adjoint operators.

The following important formula holds for the norm of a self-adjoint operator.

Theorem 14. If the operator A – is self-adjoint, then

Proof. By the Cauchy – Bunyakovsky inequality we have, for Consequently, if then

Let us prove the converse inequality. Note that any can be represented in the form where (since if then if then any vector with norm equal to one). Hence for any we have |(Az, z)| = ||z||2|(Az′, z′ )| ≤ C||z||2.

Now for any taking into account the equality we have

and, subtracting the second equality from the first, we find

Hence, and from the inequality established above |(Az, z)| ≤ C||z||2

|(Ax, y)| ≤ C(||x + y||2 + ||x – y||2)|.

Let us use the parallelogram equality (Theorem 6.8)

,

we obtain

|(Ax, y)| ≤ C(||x||2 + ||y||2)|.

Putting substitute into the last inequality . Then and we obtain or The same inequality also holds for Ax = 0. Hence, and, thereby, equality is proved.

Corollary 1. If for a self-adjoint operator for all then A=0.

Indeed, if for all then by the theorem, and hence A = 0.

For a self-adjoint operator A we further introduce the notion of its bounds – the upper and lower ones:

Corollary 2. It follows from the theorem that

From the definition of the bounds it is easily deduced that for any the relation holds

Problems

1. Are the following functionals linear in C[0, 1]?

1) ;

2) F(x)=x(1/2);

3) ;

4)

5) ;

6) ;

7) F(x)=x′(t0);

8) ;

9) ;

10) .

Which of these functionals are continuous in C[0, 1]? Compute their norms.

Which of these functionals are continuous in L2[0,1]? Compute their norms.

2. Which of the given functionals, acting on the corresponding classes of elements from l2, are linear; continuous?

1) f(x)= xksink;

2) f(x)= xk;

3) f(x)= xksgn(k-n);

4) f(x)= xk2k1/2;

5) f(x)= xkk-1/2;

6) f(x)= xk2;

7) f(x)= xk-xk-1;

8) f(x)= |xk|;

9) f(x)=supk|xk|;

10) f(x)= |xk| 2.

3. Find the norm of the functional in the space C[0, 1].

4. Are the following linear functionals continuous on the space ,

a) ;

b) ;

5. Verify that the functional

is continuous in the space ; show that the exact upper bound of its values on the closed unit ball of the space C[0,1] is equal to 1, but this upper bound is not attained at any element of the unit ball.

6. Let in a Hilbert space the sequence {xn} converge weakly to x0, i.e. (xn, y) → (x0, y) for any y ∈H, and ||xn|| → ||x0||. Show that xn → x0.

7. If in a Hilbert space the sequence {xn} converges weakly to x0 and the sequence {yn} converges in norm to y0, then (xn, yn) → (x0, y0). Is weak convergence of the sequence {yn} sufficient?

8. Prove that in a finite-dimensional space weak convergence coincides with strong convergence, i.e., convergence in norm.

See also

• Mathematical concepts
  • Hermitian operator
  • Norm (mathematics)
  • Transpose of linear maps
• Physical applications
  • Operator (physics)
  • C*-algebra