Is B(H) a Hilbert space?

If H is a Hilbert space, Is B(H) under the operator norm a Hilbert space?
If not, is there exists any norm on B(H) that makes it a Hilbert space?

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To call something a Hilbert space means that it is equipped with a complete inner product, not just a norm. A complete normed vector space is called a Banach space, and indeed $B(H)$ with the operator norm is a Banach space. Indeed, it has some additional structure (multiplication given by composition, and an involution given by taking adjoints) which makes it a C*-algebra.

Of course, you can still ask whether there exists an inner product on $B(H)$ which makes it a Hilbert space, and this depends on $H$:

If $H$ is finite-dimensional, then choosing a basis identifies it with $\mathbb{C}^n$ for $n=\dim H$, and $B(H)$ is identified with the space of $n\times n$ matrices, which is a Hilbert space.

If $H$ is infinite-dimensional, then $B(H)$ is not a Hilbert space: see this other question for some reasons why.

For your first question: we have that $B(H)$ is a Hilbert space if and only if $\dim(H) \leq 1$ holds.

  • If $\dim(H) \leq 1$ holds, then it is clear that $B(H)$ is a Hilbert space.
  • If $\dim(H) > 1$ holds, then we may choose $x,y\in H$ with $\lVert x \rVert = \lVert y \rVert = 1$ and $\langle x, y\rangle = 0$. Let $P : z \mapsto \langle z,x\rangle \cdot x$ and $Q : z \mapsto \langle z,y\rangle\cdot y$ denote the orthogonal projections onto $\text{span}(x)$ and $\text{span}(y)$, respectively, then we have
    $$ \lVert P + Q \rVert^2 + \lVert P – Q \rVert^2 \: = \: 1 + 1 \: = \: 2, $$
    but also
    $$ 2\Big(\lVert P \rVert^2 + \lVert Q\rVert^2\Big) \: = \: 2\cdot (1 + 1) \: = \: 4. $$
    We see that $B(H)$ does not satisfy the parallelogram rule, so it cannot be a Hilbert space.

As for your second question, Brad already pointed out a different norm on $B(H)$ that turns it into a Hilbert space in case $H$ is finite-dimensional. This norm can be generalised to infinite dimensions, where it is known as the Hilbert–Schmidt norm, and one can prove that this does indeed satisfy the parallelogram rule. However, unfortunately not all elements of $B(H)$ have finite Hilbert–Schmidt norm, unless $H$ is finite-dimensional. Thus, you get a proper subspace $\mathcal{HS}(H) \subsetneq B(H)$ consisting of Hilbert–Schmidt operators (that is, operators with finite Hilbert–Schmidt norm). The inner product can be given by $\langle A, B\rangle = \text{tr}(B^*A)$, where you have to prove first that the product of two Hilbert–Schmidt operators is a trace class operator.

Edit: even in the infinite-dimensional case there exists a different norm $\lVert\:\cdot\:\rVert_2$ that turns $B(H)$ into a Hilbert space, as by this answer. However, this does not provide us with an explicit representation of $\lVert\:\cdot\:\rVert_2$, and there is no guarantee that it should be a Banach algebra norm. I posted a follow-up question about this.

For more on Hilbert–Schmidt operators I know the following references (thought there may be others):

  • John B. Conway, A Course in Functional Analysis, exercises IX.2.19 and IX.2.20.
  • Gerard J. Murphy, $C^*$-algebras and operator theory, section 2.4. (Excellent book!)