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

Question

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.)

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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.

Answer 1

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.

Answer 2

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 $(\phi+\psi,\phi-\psi)$.