Identifying linear operator with a bilinear symmetric form using Theorem of Schwarz

59 Views Asked by At

We study the energy functional $E$ of the form $$E(v)=\frac{1}{2}a(v,v)+\int_{\Omega}F(x,v).$$ Let $V$ be a real Banach space with norm $||\cdot||_{V}$ and denote by $V^{'}$ the dual space. By $\langle\cdot,\cdot\rangle_{V^{'}\times V}$ we denote the duality between the space $V$ and its dual $V^{'}$.
Let $U\subset V$ be open, and let $E\in C^2(U,\mathbb{R})$. We denote by $\mathcal{M}\in C^1(U,V^{'})$ the first derivate of $E$ and by $\mathcal{L}\in C(U,B(V,V^{'}))$ the second derivate.
Let $\varphi\in U$ be a stationary point for $E$, i.e. $\mathcal{M}(\varphi)=0$.
Then, by the Theorem of Schwarz, we may identify the linear operator $\mathcal{L}(\varphi)$ with a bilinear symmetric form on $V\times V$.

I don't understand how the conclusion arises exactly. The Theorem of Schwarz yields the symmetry of the bilinear form. But why does such a bilinear form exist in the first place, given that the Riesz representation theorem only applies in Hilbert spaces?

Edit: This question is about the proof of Lemma 3.3 in https://core.ac.uk/download/pdf/82204962.pdf