1. Principles of operator algebras
  2. Linear operators
  3. 2d. Diagonal operators

2d. Diagonal operators

[math] \newcommand{\mathds}{\mathbb}[/math]

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Let us work out now what happens in the case that we are mostly interested in, namely [math]H=L^2(X)[/math], with [math]X[/math] being a measured space. We first have:

Theorem

Given a measured space [math]X[/math], consider the Hilbert space [math]H=L^2(X)[/math]. Associated to any function [math]f\in L^\infty(X)[/math] is then the multiplication operator

[[math]] T_f:H\to H\quad,\quad T_f(g)=fg [[/math]]
which is well-defined, linear and bounded, having norm as follows:

[[math]] ||T_f||=||f||_\infty [[/math]]
Moreover, the correspondence [math]f\to T_f[/math] is linear, multiplicative and involutive.


Show Proof

There are several assertions here, the idea being as follows:


(1) We must first prove that the formula in the statement, [math]T_f(g)=fg[/math], defines indeed an operator [math]H\to H[/math], which amounts in saying that we have:

[[math]] f\in L^\infty(X),\ g\in L^2(X)\implies fg\in L^2(X) [[/math]]


But this follows from the following explicit estimate:

[[math]] \begin{eqnarray*} ||fg||_2 &=&\sqrt{\int_X|f(x)|^2|g(x)|^2d\mu(x)}\\ &\leq&\sup_{x\in X}|f(x)|^2\sqrt{\int_X|g(x)|^2d\mu(x)}\\ &=&||f||_\infty||g||_2\\ & \lt &\infty \end{eqnarray*} [[/math]]


(2) Next in line, we must prove that [math]T[/math] is linear and bounded. We have:

[[math]] T_f(g+h)=T_f(g)+T_f(h)\quad,\quad T_f(\lambda g)=\lambda T_f(g) [[/math]]


As for the boundedness condition, this follows from the estimate from the proof of (1), which gives, in terms of the operator norm of [math]B(H)[/math]:

[[math]] ||T_f||\leq||f||_\infty [[/math]]


(3) Let us prove now that we have equality, [math]||T_f||=||f||_\infty[/math], in the above estimate. For this purpose, we use the well-known fact that the [math]L^\infty[/math] functions can be approximated by [math]L^2[/math] functions. Indeed, with such an approximation [math]g_n\to f[/math] we obtain:

[[math]] \begin{eqnarray*} ||fg_n||_2 &=&\sqrt{\int_X|f(x)|^2|g_n(x)|^2d\mu(x)}\\ &\simeq&\sup_{x\in X}|f(x)|^2\sqrt{\int_X|g_n(x)|^2d\mu(x)}\\ &=&||f||_\infty||g_n||_2 \end{eqnarray*} [[/math]]


Thus, with [math]n\to\infty[/math] we obtain [math]||T_f||\geq||f||_\infty[/math], which is reverse to the inequality obtained in the proof of (2), and this leads to the conclusion in the statement.


(4) Regarding now the fact that the correspondence [math]f\to T_f[/math] is indeed linear and multiplicative, the corresponding formulae are as follows, both clear:

[[math]] T_{f+h}(g)=T_f(g)+T_h(g)\quad,\quad T_{\lambda f}(g)=\lambda T_f(g) [[/math]]


(5) Finally, let us prove that the correspondence [math]f\to T_f[/math] is involutive, in the sense that it transforms the standard involution [math]f\to\bar{f}[/math] of the algebra [math]L^\infty(X)[/math] into the standard involution [math]T\to T^*[/math] of the algebra [math]B(H)[/math]. We must prove that we have:

[[math]] T_f^*=T_{\bar{f}} [[/math]]


But this follows from the following computation:

[[math]] \begin{eqnarray*} \lt T_fg,h \gt &=& \lt fg,h \gt \\ &=&\int_Xf(x)g(x)\bar{h}(x)d\mu(x)\\ &=&\int_Xg(x)f(x)\bar{h}(x)d\mu(x)\\ &=& \lt g,\bar{f}h \gt \\ &=& \lt g,T_{\bar{f}}h \gt \end{eqnarray*} [[/math]]


Indeed, since the adjoint is unique, we obtain from this [math]T_f^*=T_{\bar{f}}[/math]. Thus the correspondence [math]f\to T_f[/math] is indeed involutive, as claimed.

■

In what regards now the basic classes of operators, the above construction provides us with many new examples, which are very explicit, and are complementary to the usual finite dimensional examples that we usually have in mind, as follows:

Theorem

The multiplication operators [math]T_f(g)=fg[/math] on the Hilbert space [math]H=L^2(X)[/math] associated to the functions [math]f\in L^\infty(X)[/math] are as follows:

  • [math]T_f[/math] is unitary when [math]f:X\to\mathbb T[/math].
  • [math]T_f[/math] is a symmetry when [math]f:X\to\{-1,1\}[/math].
  • [math]T_f[/math] is a projection when [math]f=\chi_Y[/math] with [math]Y\in X[/math].
  • There are no non-unitary isometries.
  • There are no non-unitary symmetries.
  • [math]T_f[/math] is positive when [math]f:X\to\mathbb R_+[/math].
  • [math]T_f[/math] is self-adjoint when [math]f:X\to\mathbb R[/math].
  • [math]T_f[/math] is always normal, for any [math]f:X\to\mathbb C[/math].


Show Proof

All these assertions are clear from definitions, and from the various properties of the correspondence [math]f\to T_f[/math], established above, as follows:


(1) The unitarity condition [math]U^*=U^{-1}[/math] for the operator [math]T_f[/math] reads [math]\bar{f}=f^{-1}[/math], which means that we must have [math]f:X\to\mathbb T[/math], as claimed.


(2) The symmetry condition [math]S^2=1[/math] for the operator [math]T_f[/math] reads [math]f^2=1[/math], which means that we must have [math]f:X\to\{-1,1\}[/math], as claimed.


(3) The projection condition [math]P^2=P^*=P[/math] for the operator [math]T_f[/math] reads [math]f^2=f=\bar{f}[/math], which means that we must have [math]f:X\to\{0,1\}[/math], or equivalently, [math]f=\chi_Y[/math] with [math]Y\subset X[/math].


(4) A non-unitary isometry must satisfy by definition [math]U^*U=1,UU^*\neq1[/math], and for the operator [math]T_f[/math] this means that we must have [math]|f|^2=1,|f|^2\neq1[/math], which is impossible.


(5) This follows from (1) and (2), because the solutions found in (2) for the symmetry problem are included in the solutions found in (1) for the unitarity problem.


(6) The fact that [math]T_f[/math] is positive amounts in saying that we must have [math] \lt fg,g \gt \geq0[/math] for any [math]g\in L^2(X)[/math], and this is equivalent to the fact that we must have [math]f\geq0[/math], as desired.


(7) The self-adjointness condition [math]T=T^*[/math] for the operator [math]T_f[/math] reads [math]f=\bar{f}[/math], which means that we must have [math]f:X\to\mathbb R[/math], as claimed.


(8) The normality condition [math]TT^*=T^*T[/math] for the operator [math]T_f[/math] reads [math]f\bar{f}=\bar{f}f[/math], which is automatic for any function [math]f:X\to\mathbb C[/math], as claimed.

■

The above result might look quite puzzling, at a first glance, messing up our intuition with various classes of operators, coming from usual linear algebra. However, a bit of further thinking tells us that there is no contradiction, and that Theorem 2.32 in fact is very similar to what we know about the diagonal matrices. To be more precise, the diagonal matrices are unitaries precisely when their entries are in [math]\mathbb T[/math], there are no non-unitary isometries, all such matrices are normal, and so on. In order to understand all this, let us work out what happens with the correspondence [math]f\to T_f[/math], in finite dimensions. The situation here is in fact extremely simple, and illuminating, as follows:

Theorem

Assuming [math]X=\{1,\ldots,N\}[/math] with the counting measure, the embedding

[[math]] L^\infty(X)\subset B(L^2(X)) [[/math]]
constructed via multiplication operators, [math]T_f(g)=fg[/math], corresponds to the embedding

[[math]] \mathbb C^N\subset M_N(\mathbb C) [[/math]]
given by the diagonal matrices, constructed as follows:

[[math]] f\to diag(f_1,\ldots,f_N) [[/math]]
Thus, Theorem 2.32 generalizes what we know about the diagonal matrices.


Show Proof

The idea is that all this is trivial, with not a single new computation needed, modulo some algebraic thinking, of quite soft type. Let us go back indeed to Theorem 2.31 above and its proof, with the abstract measured space [math]X[/math] appearing there being now the following finite space, with its counting mesure:

[[math]] X=\{1,\ldots,N\} [[/math]]


Regarding the functions [math]f\in L^\infty(X)[/math], these are now functions as follows:

[[math]] f:\{1,\ldots,N\}\to\mathbb C [[/math]]


We can identify such a function with the corresponding vector [math](f(i))_i\in\mathbb C^N[/math], and so we conclude that our input algebra [math]L^\infty(X)[/math] is the algebra [math]\mathbb C^N[/math]:

[[math]] L^\infty(X)=\mathbb C^N [[/math]]


Regarding now the Hilbert space [math]H=L^2(X)[/math], this is equal as well to [math]\mathbb C^N[/math], and for the same reasons, namely that [math]g\in L^2(X)[/math] can be identified with the vector [math](g(i))_i\in\mathbb C^N[/math]:

[[math]] L^2(X)=\mathbb C^N [[/math]]


Observe that, due to our assumption that [math]X[/math] comes with its counting measure, the scalar product that we obtain on [math]\mathbb C^N[/math] is the usual one, without weights. Now, let us identify the operators on [math]L^2(X)=\mathbb C^N[/math] with the square matrices, in the usual way:

[[math]] B(L^2(X))=M_N(\mathbb C) [[/math]]


This was our final identification, in order to get started. Now by getting back to Theorem 2.31, the embedding [math]L^\infty(X)\subset B(L^2(X))[/math] constructed there reads:

[[math]] \mathbb C^N\subset M_N(\mathbb C) [[/math]]


But this can only be the embedding given by the diagonal matrices, so are basically done. In order to finish, however, let us understand what the operator associated to an arbitrary vector [math]f\in\mathbb C^N[/math] is. We can regard this vector as a function, [math]f(i)=f_i[/math], and so the action [math]T_f(g)=fg[/math] on the vectors of [math]L^2(X)=\mathbb C^N[/math] is by componentwise multiplication by the numbers [math]f_1,\ldots,f_N[/math]. But this is exactly the action of the diagonal matrix [math]diag(f_1,\ldots,f_N)[/math], and so we are led to the conclusion in the statement.

■

There are other things that can be said about the embedding [math]L^\infty(X)\subset B(L^2(X))[/math], a key observation here, which is elementary to prove, being the fact that the image of [math]L^\infty(X)[/math] is closed with respect to the weak topology, the one where [math]T_n\to T[/math] when [math]T_nx\to Tx[/math] for any [math]x\in H[/math]. And with this meaning that [math]L^\infty(X)[/math] is a so-called von Neumann algebra on [math]L^2(X)[/math]. We will be back to this, on numerous occasions, in what follows.

General references

Banica, Teo (2024). "Principles of operator algebras". arXiv:2208.03600 [math.OA].