# Minimalist proof that the operator norm on $H^*$ satisfies the parallelogram law?

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.

• 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 5 at 1:40
• 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 5 at 1:40
• @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 5 at 9:46
• @daw Ahhh, thanks for catching that. – Erick Wong Mar 7 at 0:13

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 $$(\phi+\psi,\phi-\psi)$$.
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.
• 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 5 at 16:11