Real-valued part of complex inner product $\DeclareMathOperator{\Re}{Re}$Let $\left<f, g\right>$ be the complex inner product defined by
$$\left<f, g\right>=\Re\left(\int_{-\infty}^{\infty}f^{*}(x)g(x)\,dx\right) $$
Could I say that this is symmetric, i.e. does $\left<f, g\right> = \left<g, f\right>$?
I know that if $f, g$ are of the form $f = a+bi$ and $g = c+di$, taking the real part of $f^{*}g$ gives
$$\Re((a-bi)(c+di)=ac+bd$$
Similarly, taking the real part of $g^{*}f$ gives
$$\Re((c-di)(a+bi))=ac+bd$$ so I know that this is symmetric.
As a function $f(x)$ can be written in the form $f(x)=a+bi$, and similarly $g(x)$ can be written in the form $g(x)=c+di$, does this mean that $$\left<f, g\right>=\Re\left(\int_{-\infty}^{\infty}f^{*}(x)g(x)\,_dx\right) = \Re\left(\int_{-\infty}^{\infty}f(x)g^{*}(x)\,dx\right) = \left<g, f\right>$$
Finally, if this is the case, and we assume that $f$ and all of its derivatives vanish at $\pm\infty$, could we say that
$$\left<f, \frac{\partial f}{\partial x}\right>=\Re\left(\int_{-\infty}^{\infty}f^{*}(x)\frac{\partial f}{\partial x}\,dx\right)=\frac{1}{2}\Re\left(\int_{-\infty}^{\infty}f^{*}(x)\frac{\partial f}{\partial x}\, dx\right) + \frac{1}{2}\Re\left(\int_{-\infty}^{\infty}f^{*}(x)\frac{\partial f}{\partial x}\,dx\right)$$
Integrating the first (or the second) by parts gives
$$\frac{1}{2}\Re\left(\int_{-\infty}^{\infty}f^{*}(x)\frac{\partial f}{\partial x}\,dx\right) + \frac{1}{2}\Re\left(\int_{-\infty}^{\infty}f^{*}(x)\frac{\partial f}{\partial x}\,dx\right)=-\frac{1}{2}\Re\left(\int_{-\infty}^{\infty}\frac{\partial f^{*}}{\partial x}f(x)\,dx\right) + \frac{1}{2}\Re\left(\int_{-\infty}^{\infty}f^{*}(x)\frac{\partial f}{\partial x}\, dx\right)$$
As this is symmetric,
$$-\frac{1}{2}\Re\left(\int_{-\infty}^{\infty}\frac{\partial f^{*}}{\partial x}f(x)\,dx\right) + \frac{1}{2}\Re\left(\int_{-\infty}^{\infty}f^{*}(x)\frac{\partial f}{\partial x}\,dx\right)\\=-\frac{1}{2}\Re\left(\int_{-\infty}^{\infty}\frac{\partial f}{\partial x}f^{*}(x)\,dx\right) + \frac{1}{2}\Re\left(\int_{-\infty}^{\infty}f^{*}(x)\frac{\partial f}{\partial x}\, dx\right)=0$$
I feel like this doesn't work, but I can't for the life of me see why. I feel like the real-valued part of the complex inner product is symmetric, but I don't know if we can apply this to the integral? Any help would be greatly appreciated, thank you!
 A: The definition you give does not define an inner product in the usual sense.  An inner product on a complex vector space $V$ must satisfy three condition for all $\alpha, \beta \in \mathbb C$ and all $u, v, w \in V$,

*

*Linearity.
$\langle u +v, w \rangle = \langle u, w \rangle + \langle v, w \rangle$ and either of $\langle \alpha u, v\rangle = \alpha \langle u,v \rangle$ or $\langle \alpha u, v\rangle = \alpha^* \langle u,v \rangle$.


*Symmetry.
$\langle u, v \rangle = \langle v, u \rangle^*$.


*Positive definite.
$\langle u, u \rangle > 0$ unless $u=0$.
In the example you give,
$$ \langle u, v \rangle = \Re \Bigg ( \int_{\mathbb R} u^*(x)v (x) dx \Bigg).$$
Here, the product is symmetric as you say, so conditions (2) holds;  moreover, condition (3) holds because $v^*\cdot v = (\Re v)^2 + (\Im v) ^2$.  However,
conditions (1) gives problems.  The first part is alright, but, for instance, if $v(x)=a(x) + ib(x)$ is any non-zero member of $V$, $\langle iv, v \rangle = 0$ (because the real part is zero) but $i\langle v, v \rangle \neq 0$ (product of $i$ and a positive real). Condition (1) requires these to be equal, so it fails with this definition.
In the second part of the question, it is not clear why you want to split the integral into two parts.  In any case, you will have,
\begin{align*}
\int_{\mathbb R} f^*(x) \frac{df}{dx}(x) ~dx = -\int_{\mathbb R} \frac{df^*}{dx}(x) f(x) ~  dx = 0
\end{align*}
which is a general property of functions that vanish at the limits $\pm\infty$.  Recall for a real valued function $g$ with this condition,
\begin{align*}
\int_{\mathbb R} g(x)g'(x) dx = \Big [ \frac{1}{2} g(x)^2 \Big]_{-\infty}^{+\infty} = 0.
\end{align*}
