$\dot{H}^1 (\mathbb R^d) = \{ f:\mathbb R^d \to \mathbb C : \nabla f \in L^{2} \}$
Then $\dot{H}^1$ is a Banach Space with respect to the norm $ \|f \|_{\dot{H}^1} = \|\nabla f \|_{L^2}.$
My Naive Questions:
(1) How to define inner product (natural) on $\dot{H}^1$? Is $\dot{H}^1$ Hilbert space? (2) How define weak convergence on $\dot{H}^1$?
Side Query: Why authors writes $\cdot$ above $H$?
My thought: I guess, $f_n$ converges to $f$ weakly in $L^2$ if $\int_{\mathbb R^d} f_n g \to \int_{\mathbb R^d} fg$ for all $g\in L^2 $ (correct me if I am wrong)
As already mentioned by Elsa, your norm is not a norm, since constant functions are mapped to zero.
The inner product has to satisfy $(f,f) = \|f\|^2$. Since $\|f\|^2 = \int_{\mathbb R^d} |\nabla f|^2 \, \mathrm{d}x$, you can use $$(f,g) = \int_{\mathbb R^d} \nabla f \cdot \nabla g \, \mathrm{d}x.$$ Alternatively, you can use the polarization identity.
By definition, in an arbitrary Banach space $X$, a sequence $\{x_n\} \subset X$ converges weakly towards $x \in X$, if $f(x_n) \to f(x)$ for all $f \in X^*$. By the Riesz representation theorem, this is equivalent to $(x_n, y)_X \to (x,y)_X$ for all $y \in X$, if $X$ is a Hilbert space.