$g_n \rightharpoonup g$ in $\dot{H}^1 \implies g_n \to g$ in $\dot{H}^1$?

Question

Let $g_n, \phi \in \dot{H}^1(\mathbb R^d)= \{f: \nabla f \in L^{2}(\mathbb R^d) \}.$

Assume that $\int_{\mathbb R^d} \nabla g_n \cdot \nabla h \ dx \to \int_{\mathbb R^d} \nabla g \cdot \nabla h \ dx$ as $n\to \infty$ for all $h\in \dot{H}^1.$ (In other words, $g_n$ converges to $g$ weakly in $\dot{H}^1$.)

Question: Can we say $ \|\nabla (g_n -g)\|_{L^2} \to 0$ as $n\to \infty$? (What is a mistake in the below solution?)

My Solution: Since $g_n \rightharpoonup g$ in $\dot{H}^1$, we have $\int \nabla (g_n-g) \cdot \nabla h \to 0$ for all $h\in \dot{H}^1.$ In particular this is true for, $h=g_n-g,$ so we have

$$\int \nabla (g_n-g) \cdot \nabla (g_n-g) dx \to 0.$$

But $\nabla (g_n-g) \cdot \nabla (g_n-g) = |\nabla (g_n -g)|^2,$ and so $\|\nabla (g_n-g)\|_{L^2}\to 0$.

Thus, $g_n \rightharpoonup g$ in $\dot{H}^1 \implies g_n \to g$ (strong convergence) in $\dot{H}^1$. I guess, this may not be true, but where does solution goes wrong?

Answer 1

Notice that you only know that $\int \nabla(g_n - g)\cdot H \to 0$ for every fixed $H \in L^2(\mathbb{R}^d,\mathbb{R}^d)$. This is different from requiring that for every sequence $\{H_n\} \subset L^2(\mathbb{R}^d,\mathbb{R}^d)$, $\int \nabla(g_n - g)\cdot H_n \to 0$, which is essentially what you are using in your proof by choosing $H_n = \nabla(g_n - g)$.