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

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?

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 $H_n = \nabla(g_n - g)$.