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Let $H$ be a complex Hilbert space. I am trying to prove that $H^*$ (the set of bounded linear maps from $H$ to $\mathbb C$) is again a Hilbert space, where the inner product induces the operator norm. I have already seen proofs that $H^*$ with the operator norm is a Banach space, and also that a norm is induced by an inner product if and only if the norm satisfies the parallelogram law

$$\|x+y\|^2+\|x-y\|^2=2\|x\|^2+2\|y\|^2.$$

So if I can show that the operator norm on $H^*$ obeys this identity, then I am done. But how can I do that?

I would prefer not to invoke the Riesz representation theorem, as I have seen some sources do. (Hence "minimalist" in the question title.)

Letting $\phi, \psi\in H^*$, we have

\begin{align} \|\phi+\psi\|^2+\|\phi-\psi\|^2 &=\left(\sup_{\|x\|=1}|\phi(x)+\psi(x)|\right)^2 + \left(\sup_{\|x\|=1}|\phi(x)-\psi(x)|\right)^2 \\\\ &=\sup_{\|x\|=1}|\phi(x)+\psi(x)|^2 + \sup_{\|x\|=1}|\phi(x)-\psi(x)|^2 \\\\ &=\sup_{\|x\|=1}\Big(|\phi(x)|^2+|\psi(x)|^2+\overline{\phi(x)}\psi(x)+\phi(x)\overline{\psi(x)}\Big)\\\\ &\qquad + \sup_{\|x\|=1}\Big(|\phi(x)|^2+|\psi(x)|^2-\overline{\phi(x)}\psi(x)-\phi(x)\overline{\psi(x)}\Big) \end{align}

If we could only combine the two "sup" expressions and cancel cross terms, and then also separate out the "sup" terms again to give $2\|\phi\|^2+2\|\psi\|^2$, we would be done. But I can't see any justification for this.

WillG
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  • I haven't thought about this in a long while, but here's an idea: you certainly get an $\ge$ inequality if you combine the sups into one expression. To get the opposite inequality, seems the map $(a,b) \to (a+b,a-b)$ is essentially a 45-degree rotation (up to scalars), so applying the same argument on $| (\phi + \psi) + (\phi - \psi) |^2 + | (\phi + \psi) - (\phi - \psi) |^2$ should get you the reverse inequality. But I could have easily missed something. – Erick Wong Mar 05 '20 at 01:40
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    I'm just going to put this out here: I don't think this approach is going to work particularly well. It's not clear how you're going to work inner products into this expression. In order to combine these terms, the only direction forward I can see involves geometry of the Hilbert Space ball, which will basically take you a stones throw from a proof of the Riesz Representation Theorem anyway. – user744868 Mar 05 '20 at 01:40
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    @ErickWong Unfortunalty, this gives $\sup_{|x|\le1} ( |\phi(x)|^2+|\psi(x)|^2)$ on the rhs, which is less than $|\phi|^2+|\psi|^2$. – daw Mar 05 '20 at 09:46
  • @daw Ahhh, thanks for catching that. – Erick Wong Mar 07 '20 at 00:13
  • Use that |ϕ(x)+ψ(x)|2+|ϕ(x)−ψ(x)|2 is always smaller than, or equal to, 2|ϕ(x)|2+2|ψ(x)|2, for all points x. Since the sup preserves order for the case when the sup is obtained by an element, you have the inequality <=. – R.W May 11 '23 at 17:46

2 Answers2

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Here is an idea, how one could prove it. Unfortunately, there is an additional factor of 2, which I cannot get rid of. Maybe someone has an idea, how to improve it.

Let $x,y\in H$. Then $$\begin{split} (\phi+\psi)(x)^2 + (\phi-\psi)(y)^2 &= (\phi(x)+\psi(x))^2 + (\phi(y)-\psi(y))^2\\ &=\frac12\Big( \phi(x+y)^2 + \phi(x-y)^2 + \psi(x+y)^2 + \psi(x-y)^2 \\ & \qquad+ 2\phi(x+y)\psi(x-y) + 2\phi(x-y)\psi(x+y) \Big)\\ &\le \frac12\Big( (\|\phi\|^2+\|\psi\|^2)(\|x+y\|^2+\|x-y\|^2) + 4\|\phi\|\|\psi\|\|x+y\|\|x-y\|\Big)\\ & \le (\|\phi\|^2+\|\psi\|^2)(\|x+y\|^2+\|x-y\|^2)\\ & = 2(\|\phi\|^2+\|\psi\|^2)(\|x\|^2+\|y\|^2). \end{split}$$ First inequality is Cauchy-Schwarz, second inequality uses $2ab\le a^2+b^2$ twice. In the last step, I used the parallelogram identity in $H$. Taking the supremum over $x$ and $y$ with $\|x\|\le1$, $\|y\|\le1$ of this inequality, yields $$ \|\phi+\psi\|^2 + \|\phi-\psi\|^2 \le 4(\|\phi\|^2+\|\psi\|^2). $$ If we could prove this inequality with factor $2$ instead of $4$, we could finish, by substituting $(\phi,\psi)$ by $(\phi+\psi,\phi-\psi)$.

daw
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  • This seems like a promising route, but I can't see a way to eliminate the factor of 2 either. – WillG Mar 05 '20 at 15:48
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By Riesz's theorem, there is an antilinear isometric bijection $I:H^\ast\to H$ such that $\varphi(x)=(x,I\varphi)$. Antilinear means that $I(\alpha\varphi_1+\beta\varphi_2)=\overline{\alpha}I(\varphi_1)+\overline{\beta}I(\varphi_2)$. As you know, $H^\ast$ is Banach space. We define a function $<\varphi_1,\varphi_2>=(I\varphi_2,I\varphi_1)$ in it. All you need to do is show that $<\cdot,\cdot>$ -- is inner product in $H^\ast$ and $<\varphi,\varphi>=\|\varphi\|^2$. It will follow from here that the space $H^\ast$ is Hilbert and the parallelogram law will turn out automatically.

thing
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    Isn’t this exactly the type of argument the OP wants to avoid? – Erick Wong Mar 05 '20 at 05:44
  • You can’t think of anything more than “minimal”. No one asks to prove the Riesz theorem, it is known. – thing Mar 05 '20 at 09:39
  • This seems like a valid proof, but I am looking for something that doesn't involve Riesz's theorem. It is also possible that the norm induced by the inner product from Riesz's theorem is different from the operator norm, so this would also need to be shown. My hope was to avoid all that if possible. – WillG Mar 05 '20 at 15:48
  • The norm induced by the introduced inner product is the operator norm. I wrote that $<\varphi,\varphi>=|\varphi|^2$, and it's very easy to proof. In fact, this follows from $|I\varphi|=|\varphi|$. – thing Mar 05 '20 at 16:11