This question arose from my previous question. I want to prove the following inequality. Let $f \in H^{1}(\mathbb{R}^{d})$ and $p^{*} = 2d/(d-2)$. Then: $$ \|f\|_{L^{p^{*}}} \le C\|\nabla f\|_{L^{p^{*}}}$$ for some constant $C > 0$ which is independent of $f$. In the comments of my previous question it was pointed out that this inequality is related to the Gagliardo-Nirenberg-Sobolev inequality, which states that (see, e.g. Evans): $$\|f\|_{L^{p^{*}}} \le C \|\nabla f\|_{L^{p}}$$ for all $1 \le p < d$ and $p^{*} = dn/(n-p)$ and $f \in C_{c}^{\infty}(\mathbb{R}^{d})$. As you can see, the later reduces to the first one when $p=2$, so $p^{*} = 2d/(d-2)$; the only important difference is that the second one is an inequality for $C_{c}^{\infty}(\mathbb{R}^{d})$ functions, whilst the first one is for $H^{1}(\mathbb{R}^{d})$. However, I was thinking about using density arguments to prove the first one from the second one. Let me sketch the argument below.
Disclaimer: I don't want to use any of the Sobolev inequalities in the following.
Fact 1: $C_{c}^{\infty}(\mathbb{R}^{d})$ is dense in $L^{p}(\mathbb{R}^{d})$ for every $1 \le p \le \infty$.
Fact 2: $C_{c}^{\infty}(\mathbb{R}^{d})$ is dense in $H^{1}(\mathbb{R}^{d})$ (I mean in the topology of $H^{1}(\mathbb{R}^{d})$).
Now, norms are continuous functions, so I wonder if the following can be used. Let $f \in H^{1}(\mathbb{R}^{d})$ so that, in particular, $f \in L^{2}(\mathbb{R}^{d})$. Let $\{\varphi_{n}\}_{n\in \mathbb{N}}\subset C_{c}^{\infty}(\mathbb{R}^{d})$ such that $\varphi_{n} \to f$ in $H^{1}(\mathbb{R}^{d})$. Then, $\varphi_{n} \to f$ in $L^{2}(\mathbb{R}^{d})$ and $\nabla \varphi_{n} \to \nabla f$ in $L^{2}(\mathbb{R}^{d})$ (these facts I already proved and can be used).
The problem is:
First: why $f \in L^{p^{*}}(\mathbb{R}^{d})$, with $p^{*}= 2d/(d-2)$ in the first place, so I can take the norm $\|f\|_{L^{p^{*}}}$? Second: does the convergence $\varphi_{n} \to f$ in $L^{2}$ imply convergence in $L^{p^{*}}$ somehow? My idea then would be to use the Gagliardo-Nirenberg-Sobolev inequality to get: $$\lim_{n\to \infty}\|\varphi_{n}\|_{L^{p^{*}}} = \|f\|_{L^{p^{*}}} \le C \lim_{n\to \infty}\|\nabla \varphi_{n}\|_{L^{2}} = C\|\nabla f\|_{L^{2}}.$$
Does this idea work somehow?
First: Fatou’s lemma is the key. Whenever you have convergence in $L^2$, you have convergence a.e. (up to subsequences). And whenever you have a sequence of functions $\varphi_n$ that converge a.e. to a function $f$, and you have a uniform bound on the $L^{p^*}$ norms, then the limit is also in $L^{p^*}$ with the same bound. The uniform bound on $\|\varphi_n\|_{L^{p^*}}$ is guaranteed by the Sobolev inequality for test functions, since the $\varphi_n$ are bounded in $H^1$. This is actually all you need to prove the inequality, you don’t need the convergence $L^{p^*}$.
Second. The convergence in $L^{p^*}$ of course follows a posteriori after you prove the estimate for all $H^1$ functions, but you can actually show that beforehand: $\varphi_n$ is a cauchy sequence in $H^1$, and by the Sobolev estimate for test functions (and by linearity) it follows immediately that it is a Cauchy sequence in $L^{p^*}$, so it converges to some function $g$ in $L^{p^*}$. Also, it converges to $f$ in $L^2$, and since the two convergences are also a.e. (up to subsequences), and the a.e. limit is unique, it follows that $f=g$ a.e. If you want, this is an alternative proof to the one I wrote in the first point.
More generally, this kind of inequalities extend to all functions in any given Banach space $X$ for which the right-hand side is bounded by the norm of $X$, and such that test functions are dense in $X$. And you can also prove that $X$ is a subspace of the space defined by the norm on the left-hand side using real analysis tricks like Fatou’s lemma, or other characterization properties for more general Banach spaces.
Extending a linear estimate by density to a larger space is extremely common and standard, so that nobody ever writes down the details. Whenever you see an estimate for test functions (or for functions on a very small space that is not related to the norm on the right-hand side), you should know that the estimate extends by density to a much larger space. In fact, the Sobolev estimate is often stated only for test functions, but everyone knows that it extends to all functions in $W^{1,p}$, or (for $p<d$) to $\dot W^{1,p}$.