A NOTE ON THE DEFINITION OF AN UNBOUNDED NORMAL OPERATOR
MATTHEW ZIEMKE
1. Introduction
Let H be a complex Hilbert space and let T : D(T )(⊆ H) → H be a generally unbounded linear operator. In order to begin the discussion about whether or not the operator T is normal, we need to require that T have a dense domain. This ensures that the adjoint operator T ∗ is defined. In the literature, one sees two definitions for what it means for a densely defined operator T to be normal. The first requires that T be closed and that T ∗T = TT ∗ while the second requires that D(T ) = D(T ∗) and kT xk = kT ∗xk for all x ∈ D(T ) = D(T ∗). There is no harm in having these competing definitions as they are, in fact, equivalent. While the equivalence of these statements seems to be widely known, the proof is not obvious, and my efforts to find a detailed and self-contained proof of this fact have left me empty-handed. I attempt to give such a proof here. I also want to note that while we say a bounded linear operator T is normal if and only if T ∗T = TT ∗, the lone requirement of T ∗T = TT ∗ for an unbounded densely defined operator does not, in general, imply T is normal. I used [1] and [2] to write this note.
2. Proof of the equivalence
We begin with a definition.
Definition 2.1. Let T : D(T )(⊆ H) → H be a densely defined operator. We define the
T -inner product on D(T ) × D(T ), denoted as h·, ·iT , by the equation
hx, yiT = hx, yi + hT x, T yi, for all x, y ∈ D(T ).
It is easy to see that the T -inner product is, in fact, an inner product on D(T ). Further, 1/2 we denote the T -norm on D(T ) as k · kT and it is given by kxkT = hx, xiT . Note that
1/2 1/2 2 21/2 kxkT = hx, xiT = (hx, xi + hT x, T xi) = kxk + kT xk . 1 ZIEMKE
The following lemma is fairly elementary and widely known, but for the sake of having the notes self-contained we give the statement and proof.
Lemma 2.2. Let T : D(T )(⊆ H) → H be a densely defined operator. The inner product
space (D(T ), k · kT ) is a Hilbert space if and only if T is closed.
Proof. First, assume T is closed. Let (xn)n≥1 ⊆ D(T ) be a Cauchy sequence with respect to the T -norm. Then, for n, m ∈ N, we have that
2 2 2 kxn − xmk + kT xn − T xmk = kxn − xnkT → 0 as m, n → ∞ and so (xn)n≥1 and (T xn)n≥1 are Cauchy with respect to the norm k · k. So, there exists k·k k·k x ∈ H and y ∈ H such that xn −→ x and T xn −→ y. Since T is closed, x ∈ D(T ) and
T xn → T x. Further,