$\dot{H}^1 (\mathbb R^d) = \{ f:\mathbb R^d \to \mathbb C : \nabla f \in L^{2} \}$

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)

  • 1
    $\begingroup$ I am not sure whether that is really a norm. Gradiant could be zero yielding a zero for the "norm" even though $f$ might not be zero... $\endgroup$
    – Elsa
    Feb 21, 2017 at 4:48

1 Answer 1


As already mentioned by Elsa, your norm is not a norm, since constant functions are mapped to zero.

  1. 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.

  2. 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.

  • $\begingroup$ @grew: Thanks. But in innner product space also has to satisfies $ (x, x) =0$ iff $x=0$. (And so the function 1 creates problem?) $\endgroup$
    – abcd
    Feb 21, 2017 at 9:14
  • $\begingroup$ @grew: If we define completion of $C_c^{\infty}(\mathbb R^d)$ with respect to the norm $\|u\| = \| \nabla u \|_{L^2}$. Is this completion same as $\dot{H}^{1}$? Thanks $\endgroup$
    – abcd
    Feb 21, 2017 at 9:16
  • $\begingroup$ Concerning your second comment: $\|\cdot\|$ is not a norm. For defining the completion, you need a norm. $\endgroup$
    – gerw
    Feb 21, 2017 at 10:12
  • $\begingroup$ @gerw Actually, it is a norm on $C_c^\infty$. Constant functions are ruled out. $\endgroup$ Feb 26, 2017 at 16:24

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