# $g_n \rightharpoonup g \implies T_{\phi}(g_n) \to T_{\phi} (g)$ ($T_\phi$ is bounded functional)

Put $\dot{H}^1= \{f: \nabla f \in L^2\}.$

Choose $\phi \in \mathcal{S}(\mathbb R^d)$ (Schwartz space), and define $T_{\phi}:\dot{H}^1(\mathbb R^d) \to \mathbb C$ as $T_\phi (f)= \int_{\mathbb R^d} f(x) \phi (x) dx$, so that $T_\phi$ is a bounded linear function on $\dot{H}^1(\mathbb R^d),$ (that is, $|T_{\phi}(f)| \leq C \|\nabla f\|_{L^2},$ where $C$ is constant)

Assume that $g_n \rightharpoonup g$ in $\dot{H}^1$, that is $\int \nabla g_n \cdot \nabla h \to \int \nabla g \cdot \nabla h$ for all $h\in \dot{H}^1.$

Question: Can we expect that $T_{\phi}(g_n) \to T_{\phi} (g)$ in $\mathbb C$?

• Yes of course $h = \phi$ Mar 2, 2017 at 5:55
• @user1952009: Thanks. I have edited question. Is it still true if replaced $L^2$ by $\dot{H}^1$?
– abcd
Mar 2, 2017 at 6:27
• $g_n(x) = n$... Mar 2, 2017 at 9:45
• @user1952009: Thanks, and sorry for the mess-up. I should have define $\dot{H}^1$ by taking factoring out constants. See answer to this question
– abcd
Mar 2, 2017 at 10:05
• And then can we expect?
– abcd
Mar 2, 2017 at 10:06

Yes, you can use the Riesz representation theorem to get $h \in \dot H^1$ with $$T_\phi(g) = \int \nabla g \cdot \nabla h$$ for all $g \in \dot H^1$.
• Thanks. One side question: How I should make $\dot{H}^1$ Hilbert space? How should I take the factor by constants, etc. , so everything make sense? Can tell you me bit on this?
• @abcd No because as the OP defined it, it is not a Hilbert space. Take $g_n(x) = n$. Now if $\lim_{x \to \infty} g_n(x) = 0$ then yes Mar 2, 2017 at 11:11
• @user1952009: Thanks. I understand constants creates a problem. But can we expect to make it a Hilbert space by removing (factoring) constants? Also OP mentioned that it is a subspace of $\mathcal{S}'/ \mathbb C$ (I do not know how to justify this...)
• @abcd I just did : $\lim_{x \to \infty}g_n(x) = 0$. Mar 2, 2017 at 11:23
• @abcd if $d = 1$, can you find $h$ ? Mar 2, 2017 at 11:25