The set of all self-adjoint operators are closed in Hilbert space.

Let $T \in B(H)$ where $H$ is the Hilbert space and $B(H)$ denotes the set of linear operators from $H \mapsto H$.

We are working with the operator norm.

Adjoint of an operator $T$ is denoted by $T^*$.

An operator $T$ is said to self-adjoint if $T = T^*$ where $T^*$ is the adjoint of operator $T$.

To prove the above statement let us consider a sequence of self - adjoint operators $\{T_{n}\}$ converging to a operator $T$ , now if we prove that the operator $T$ is self-adjoint then we would prove that the set of all self-adjoint operators is closed.

So $T_{n} \rightarrow T$

To prove that $T$ is self-adjoint

Proof -

Let us consider $\|T -T^*\| = \|T - T_{n} + T_{n} - T_{n}^* + T_{n}^* - T^*\|$

By triangle inequality we have $\|T -T^*\| \leq \|T -T_{n}\| + \|T_{n} - T_{n}^*\| + \|T_{n}^* - T^*\|$

So we have now that $T_{n} = T_{n}^*$ as $T_{n}$ are self-adjoint operators so the middle term become zero.

so we are left with $\|T -T^*\|\leq \|T -T_{n}\| + \|T_{n}^* - T^*\|$.

Now the term $\|T - T_{n}\| \rightarrow 0$ but how to deal with the second term $\|T_{n}^* - T^*\| = \|T_{n} - T^*\|$?

As if we showed the RHS to be zero then $T= T^*$ implying $T$ is self-adjoint and hence the set of all self-adjoint operators are closed!

  • 1
    $\begingroup$ Can you please clarify the question? What do you mean by "adjoint operator"? It looks like you are actually studying self-adjoint operators. Also, when you say closed, are you talking about operator norm topology? Are you working on a Banach space, Hilbert space, something else? $\endgroup$ Commented Nov 2, 2017 at 13:40
  • $\begingroup$ Yes,nice catch just did an edit! $\endgroup$
    – BAYMAX
    Commented Nov 2, 2017 at 13:46
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    $\begingroup$ An important equality is: $\|B \| = \|B^*\|$, for any bounded operator $B$ over a Hilbert space. $\endgroup$
    – Ranc
    Commented Nov 2, 2017 at 13:57

2 Answers 2


From the help of Functional analysis chatroom chatroom, here.

We know that the map $S \mapsto S^*$ is an isometry so $\|S\| = \|S^*\|$ where $S$ is a bounded linear operator, so applying to $S = T_{n}^* - T^* $ then we get $\|T_{n}^* - T^*\| = \|T_n - T\|$ .

So in the last term we get, $\|T -T^*\| \leq 2 \|T_{n} - T\| \rightarrow 0$.

So $T = T^*$


A bounded operator $T$ is selfadjoint iff $\langle Tx,y\rangle = \langle x,Ty\rangle$ holds for all $x,y\in H$. If $\{ T_n \}$ is a sequence of selfadjoint operators that converges in the uniform, strong, or weak sense to a bounded operator $T$, then $T$ must be selfadjoint.

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    $\begingroup$ Or, stated another way: For fixed $x,y$, the set of $T$ satisifying $\langle Tx,y\rangle = \langle x,Ty\rangle$ is a closed set. The set of self-adjoint operators is the intersection of all these closed sets, so is itself closed. $\endgroup$
    – GEdgar
    Commented Nov 2, 2017 at 14:26

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