Let $\mathbb{D}^n \subseteq \mathbb{R}^n$ be the closed $n$-dimensional unit ball, and let $A:\mathbb{D}^n \to \mathbb{R}^k \otimes \mathbb{R}^k$ be smooth. Suppose that $ \langle A , V \otimes V \rangle_{L^2} \ge 0$ for every smooth map $V:\mathbb{D}^n \to \mathbb{R}^k$.

Explicitly, I assume that $ \int_{\mathbb{D}^n} \langle A(x) , V(x) \otimes V(x) \rangle_{\mathbb{R}^k \otimes \mathbb{R}^k} dx \ge 0$ where $\langle , \rangle_{\mathbb{R}^k \otimes \mathbb{R}^k}$ is the tensor product metric on $\mathbb{R}^k$, i.e.

$$ \langle v_1 \otimes v_2 , w_1 \otimes w_2\rangle := \langle v_1, w_1 \rangle_{\mathbb{R}^k} \cdot \langle v_2 ,w_2 \rangle_{\mathbb{R}^k}.$$

Question: Is it true that $ \langle A(x) , v \otimes v \rangle_{\mathbb{R}^k \otimes \mathbb{R}^k} \ge 0$ for every $x \in \mathbb{D}^n$ and every $v \in \mathbb{R}^k$?

This seems like a "one-sided" analogue of the fundamental lemma of the calculus of variations.


Well, after consulting with a colleague, the answer seems to be positive.

Fix some point $x_0$ in the interior of $\mathbb{D}^n$, and some $v \in \mathbb{R}^k$.

Set $V_{\epsilon}(x)=d_{\epsilon}(x_0-x)v$, where $d_{\epsilon}^2=\eta_{\epsilon}$ is the standard mollifier which converges to the "delta function". Then, we have

$$\int_{\mathbb{D}^n} \langle A(x) , V_{\epsilon}(x) \otimes V_{\epsilon}(x) \rangle_{\mathbb{R}^k \otimes \mathbb{R}^k} dx=\int_{\mathbb{D}^n} \eta_{\epsilon}(x_0-x) \langle A(x) , v \otimes v \rangle_{\mathbb{R}^k \otimes \mathbb{R}^k} dx= \int_{\mathbb{D}^n} \eta_{\epsilon}(x_0-x) f(x) dx=(\eta_{\epsilon} * f)(x_0), $$ where $f(x)=\langle A(x) , v \otimes v \rangle_{\mathbb{R}^k \otimes \mathbb{R}^k}$.

Thus, we obtain $$ f(x_0)=\lim_{\epsilon \to 0}(\eta_{\epsilon} * f)(x_0) =\lim_{\epsilon \to 0} \int_{\mathbb{D}^n} \langle A(x) , V_{\epsilon}(x) \otimes V_{\epsilon}(x) \rangle_{\mathbb{R}^k \otimes \mathbb{R}^k} dx$$ which is non-negative by our assumption.


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