Proving that, if $\Vert f\Vert_{H^1}^2-\Vert g\Vert_{H^1}^2\to0$ and $\big\vert F(f)-F(g)\big\vert\to0$, then $\Vert f-g\Vert_{H^1}\to0$

Question

Motivated by a comment on a paper I was trying to prove the following (basic?) property. From now on let us denote by $X:=H^1(\mathbb{R})\cap W^{1,4}(\mathbb{R})$, where $H^1$ and $W^{1,4}$ denote the standard Sobolev spaces. Define the functional: $$ F(f):=\int_{\mathbb{R}} \big(f^4+2f^2(f')^2-\tfrac{1}{3}(f')^4\big)dx. $$ I am wondering if the following property is true: Let $g\in X$ and consider any sequence of functions $\{f_n\}\subset X$ such that $$ \big\vert\Vert f_n\Vert_{H^1}^2-\Vert g\Vert_{H^1}^2\big\vert+\big\vert F(f_n)-F(g)\big\vert\to0 \quad \hbox{as}\quad n\to+\infty. $$ Then, $\Vert f_n-g\Vert_{H^1}\to 0$ as $n\to+\infty$. I have the feeling that this should be true. I was wondering also, what if I change $F$ by the following functional (for example): $$ F(f):=\int_{\mathbb{R}}\big(f^3+f(f')^2\big)dx. $$ Does the same property hold in this case?

Answer 1

If you're only comparing global quantities ($\| g \|_{H^1}$ or $F(g)$), there is no hope for that.

Example: take $g \in C_c^1(\mathbb{R})$ to be any non-zero $C^1$ function with compact support, and take $f_n(x) := g(x-n)$. Then $\| f_n \|_{H^1} = \| g \|_{H^1}$ and $F(f_n) = F(g)$ for each $n$, but $f_n$ does not converge to $g$.