From my limited experience with inner product spaces, it seems like the inner product being linear in the second argument would facilitate smoother notation. For instance, for $x \in H$, we could define $x^* \in H^*$ by $$x^*y = \langle x, y\rangle $$ Then this would generalize the fact that $x^T y = \langle x, y\rangle$ on $\mathbb{R}^n $.
Does linearity in the first argument make for smoother notation in some other aspect of Hilbert space theory?
Best Answer
I have taught linear algebra using both conventions and I agree with your conclusion. I found the "physicist" convention having more advantages than disadvantages when working over $\mathbb{C}$ (or working simultaneously over $\mathbb{F}$ where $\mathbb{F} \in \left \{ \mathbb{R}, \mathbb{C} \right \}$). Those include:
The only mildly annoying thing I noticed with the "physicist" convention is that the defining property for the adjoint operator is naturally written as $\left< T^{*}v, w \right> = \left< v, Tw \right>$ while I was used to the form $\left< Tv, w \right> = \left< v, T^{*}w \right>$. Both forms are equivalent but if one wants to use the Riesz anti-isomorphism to justify the existence of $T^{*}$, the form $\left< T^{*}v, w \right> = \left< v, Tw \right>$ is more natural and takes some time getting used to.