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Visualization can only get you so far. While I'm not aware of a convenient way to visualize dual spaces, the theorem you mention may be conceptualized by saying that it exhibits a symmetry of the pairing of a vector space with its dual: not only can an evil element of the dual eat a vector to yield a scalar, but the vector itself can turn into a dual-vector ...


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After having several thoughts, understood that the vector and bias can be adjusted so that the product can be fit to 1. Not 100% sure this is the correct understanding. Appreciate feedback.


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For the map $\Phi \colon H \to U'$, $x \mapsto \langle \cdot, x \rangle$ we have $\ker \Phi = U^\perp$. Hence $\Phi$ is injective if and only if $U^\perp = 0$. Because $H = U^\perp \oplus \overline{U}$ we have $U^\perp = 0$ if and only if $\overline{U} = H$, i.e. if $U$ is dense.


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The dual polyhedron corresponds to the dual set. In fact, for any set $K \subset \mathbb R^n$, you have $K^\circ = (\operatorname{conv}(K \cup \{0\}))^\circ$, where $\operatorname{conv}$ denotes the convex hull. Now, let $P = \operatorname{conv}(\{p_1, \ldots, p_N\})$ be a polyhedron. Then, you have \begin{align}P^\circ &= \{p_1,\ldots, p_N\}^\circ = ...


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One can define $\Delta$ in $H^1$ because $\Delta$ is in divergence form $$\Delta u = \text{div}\nabla u$$ Thus we define $\Delta : H^1_0(\Omega) \to H^1_0(\Omega)^*$ by $$\Delta f (\phi) := -\int_\Omega \nabla f \cdot \nabla \phi.$$ The right hand side is well defined for $f , \phi \in H^1_0(\Omega)$. If $f$ is in $H^2_0$, then integration by part gives ...


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Can linear duals (i.e. linear functionals) be represented using the geometric algebra formalism? Yes and no. In geometric algebra, dual vectors can be computed through Hodge duality. Let $\{u_1, u_2, \ldots, u_n\}$ be an orthogonal basis set for an $n$-dimensional vector space. Let $I$ be their geometric product, which is grade-$n$ due to orthogonality. ...


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Two comments: When you find the KKT point for the Lagrangian minimization - how do you know that it is actually the minimizer? A KKT point is a generalization of a stationary point in unconstrained minimization, and those as we know can be minima, maxima or saddle points. (Hint: the problem is convex). When you are done maximizing the dual function - how ...


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If you use $(u, v)_{H_0^1} = (\nabla u, \nabla v)_{L^2}$, you get $R = -\Delta$, as you have shown. And there is another scalar product on $H_0^1$, such that $R = I - \Delta$. Which one?



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