Is this subset of the Sobolev space $W^{1,q}$ closed?

Question

Consider the subset of $W^{1,q}(\mathbb{R}^n)$:
$$ A = \{f \in W^{1,q}(\mathbb{R}^n); \,\, \|f\|_{q} \leq R_1, \,\, \|\nabla f\|_{q} \leq R_2 \} $$
where $R_1 > R_2 > 0$ and the norm of $W^{1,q}(\mathbb{R}^n)$ is defined as $\|f\|_{1,q} = \max \left\{ \|f\|_{q}, \|\nabla f\|_{q} \right\}.$ I want to show that $A$ is a closed subset of $W^{1,q}(\mathbb{R}^n)$. For this, I will consider a sequence of functions $(f_n) \in A$ converging to $f \in W^{1,q}(\mathbb{R}^n)$. Let us prove that $f \in A$. Here, I am a bit uncertain about how to involve $R_1$ and $R_2$. The Dominated Convergence Theorem does not seem to work because the bounds are given by constants. I thought about using the triangle inequality $\|f\|_q \leq \|f_n - f\|_q + \|f_n\|_q \leq R_1 + \varepsilon$, and similarly for the gradient. Is this correct?

Answer 1

Your method will work, but you will just have to combine the inequalities for the function and its weak derivative in the correct way. I will give a different proof, the ideas of which you could adapt to make your argument work as well. Fix some $f\in W^{1,q}\backslash A$. By definition, either $\|f\|_q> R_1$ or $\|\nabla f\|_q>R_2$. Thus we can choose $$\varepsilon =\min\left\{\max\{\|f\|_q-R_1,0\},\max\{\|\nabla f\|_q-R_2,0\}\right\}>0.$$ Now observe that if $g\in B(f;\varepsilon)$, then $$\|g\|_q\geq\|f\|_q-\|f-g\|_q\geq \|f\|_q-\|f-g\|_{1,q}>||f||_q-\varepsilon,$$ and similarly $$\|\nabla g\|_q\geq\|\nabla f\|_q-\|\nabla(f-g)\|_q\geq \|\nabla f\|_q-\|f-g\|_{1,q}>||\nabla f||_q-\varepsilon.$$ So, by the definition of $\varepsilon$, we must have that at least one of $\|g\|_q> R_1$ or $\|\nabla g\|_q>R_2$ is true. Therefore $B(f;\varepsilon)\cap A=\emptyset$, so $f$ is an interior point of $W^{1,q}\backslash A$. As $f$ is arbitrary, we conclude that $W^{1,q}\backslash A$ is open, so indeed $A$ is closed.

Answer 2

$f_n \to f$ in $W^{1, q}$ implies $f_n \to f$ in $L^q$ and $\nabla f_n \to \nabla f$ in $L^q$. Norms are continuous by the triangle inequality, so $\|f_n\|_q \to \|f\|_q$ and $\|\nabla f_n\|_q \to \|\nabla f\|_q$.

Answer 3

Let's proceed with your sequence $(f_n)_n \subset A$. $f_n \to f$ in $W^{1,q}$ is equivalent to $$\|f_n - f\|_{1,q} = \max(\|f_n - f\|_q, \|\nabla f_n - \nabla f\|_q) \xrightarrow[n \to \infty]{} 0$$ Now, it is easy to see that if $\max(a_n, b_n) \to 0$ for non-negative sequences $(a_n)_n$ and $(b_n)_n$, then we have both $a_n \to 0$ and $b_n \to 0$.
Therefore: $$\|f_n - f\|_q \to 0,\quad \|\nabla f_n - \nabla f\| \to 0$$ This implies that, for all $\varepsilon > 0$ and $N(\varepsilon)$ such that $\|f_{N(\varepsilon)} - f\|_q \leq \varepsilon$: $$\|f\|_q \leq \|f_{N(\varepsilon)}\|_q + \|f - f_{N(\varepsilon)}\|_q \leq R_1 + \varepsilon$$ This being true for all $\varepsilon > 0$ yields: $$\|f\|_q \leq R_1$$ This applies in a simliar fashion to the gradients, and so $A$ is indeed closed.