Is $\liminf_{\epsilon \to 0} \frac{1}{\epsilon}\int_0^T \int_\Omega \nabla F(v)\nabla (T_\epsilon(u-v)) \geq 0$?

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With $\Omega$ a bounded domain, let $$A(\epsilon) = \frac{1}{\epsilon}\int_0^T \int_\Omega \nabla F(v)\nabla (T_\epsilon(u-v))$$ where $T_\epsilon$ is the truncation at height $\epsilon$: $$T_\epsilon(x) = \begin{cases} -\epsilon &: x \in (-\infty, -\epsilon]\\ x & x \in (-\epsilon, \epsilon)\\ \epsilon &: x \in [\epsilon, \infty) \end{cases},$$ $u$ and $v$ belong to $L^2(0,T;H^1(\Omega))$ and $F$ is smooth, Lipschitz, monotone and $F(0)=0$ with derivative bounded away from zero and infinity.

Question is: can we obtain $\liminf_{\epsilon \to 0} A(\epsilon) \geq 0$?

By the chain rule and Stampaachia theorem, we can see that pointwise the integrand does tend to zero. Unfortunately DCT doesn't apply thanks to the factor of $\epsilon^{-1}$.

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The following produces a counterexample.

Set $\Omega = (0,1)$, $-F(v)(x) = (u-v)(x) = x$ (and neglect the dependence on time). Then, $$A(\epsilon) = \frac1\epsilon \int_0^\varepsilon -1 \cdot 1 \equiv -1.$$