Let $\lambda$ denote the Lebesgue measure on $\mathcal B(\mathbb R)$, $T>0$, $I:=(0,T)$, $X,Y$ be $\mathbb R$-Banach spaces and $\iota$ be a continuous embedding of $X$ into $Y$.
If $p\in[1,\infty]$, say $f\in\mathcal L^1(I,X)$ has a weak derivative in $L^p(I,Y)$ if there is a $g\in\mathcal L^p(I,Y)$ with $$\int_I\varphi'\iota f\:{\rm d}\lambda=-\int_I\varphi g\:{\rm d}\lambda\;\;\;\text{for all }\varphi\in C_c^\infty(I).$$ In that case, $f':=g$.
If $p,q\in[1,\infty]$, let $E^{p,\:q}(I)$ denote the space of all $f\in\mathcal L^p(I,X)$ which admit a weak derivative in $L^q(I,Y)$. Moreover, let $$\left\|f\right\|_{E^{p,\:q}}:=\left\|f\right\|_{L^p(I,\:X)}+\left\|f'\right\|_{L^p(I,\:Y)}\;\;\;\text{for }f\in E^{p,\:q}.$$
How can we show that if $p,q\in[1,\infty)$, then $C^\infty(I,X)$ is a dense subset of $E^{p,\:q}$?
There is a proof of this claim in Mathematical Tools for the Study of the Incompressible Navier-Stokes Equations and Related Models (Lemma II.5.10):
The mentioned corollary is the (easy to verify) fact that $$\forall r\in[1,\infty):\forall g\in L^r(I,X):\left\|g\circ\tau_h-g\right\|_{L^r(I,\:X)}\xrightarrow{h\to\infty}0\tag2,$$ where $$\tau_h:\mathbb R\to\mathbb R\;,\;\;\;a\mapsto a+h$$ for $h\in\mathbb R$.
I don't understand the argumentation in the proof. How can we give a rigorous proof? And can we even find a suitable subspace "$E_0^{p,\:q}(I)$" of $E^{p,\:q}$ s.t. $C_{\color{red}c}^\infty(I,X)$ is dense in $E_0^{p,\:q}(I)$?
Let $f\in E^{p,\:q}(I)$. We can clearly find $0\le\theta_i\in C_c^\infty(\mathbb R)$ (note that I'm not taking $\theta_i\in C^\infty(I)$ as in the book excerpt, since I think we need to $\theta_i$ to be defined on all of $\mathbb R$ in what follows, but please let me know in the comments if I'm missing something) with $\theta_1+\theta_2=1$ and $\operatorname{supp}\theta_1\subseteq\left[0,\frac23T\right]$, $\operatorname{supp}\theta_2\subseteq\left[\frac13T,T\right]$ (although I'm not sure why exactly we need to consider these $\theta_i$).
Now let $$g:=\begin{cases}\theta_1f&\text{on }I\\0&\text{on }\mathbb R\setminus I.\end{cases}$$ I'm not sure whether this will be important, but we should be able to show that $g\in E^{p,\:q}(\mathbb R)$ and $g'=\theta_1f'+\theta_1'\iota f$. Now let $\eta$ be a "mollifying kernel", i.e.
- $\eta\in C_c^\infty(\mathbb R)$ with $\operatorname{supp}\eta\subseteq(-1,1)$;
- $\eta\ge0$ and $\int_{-\infty}^\infty\eta=1$;
- $\forall x,y\in\mathbb R:|x|=|y|\Rightarrow\eta(x)=\eta(y),$
and set $$\eta_\varepsilon(x):=\frac1\varepsilon\eta\left(\frac x\varepsilon\right)\;\;\;\text{for }x\in\mathbb R^d.$$ Note that $$g_h:=g\circ\tau_h\in L^p(\mathbb R)$$ and $$g_{h,\:\varepsilon}:=g_h\ast\eta_\varepsilon\in C^\infty(\mathbb R)$$ for all $h\in\mathbb R$ and $\varepsilon>0$.
Now, if I got the comment from daw in the comments below right, the idea is to show that $$\left\|g_{h,\:\varepsilon}-g\right\|_{E^{p,\:q}}\le\left\|g_h-g\right\|_{E^{p,\:q}}+\left\|g_{h,\:\varepsilon}-g_h\right\|_{E^{p,\:q}}\to0;\tag1$$ most probably as $h$ and $\varepsilon$ tend to $0$ (right?).
However, the only things which are clear to me are $$\left\|g_h-g\right\|_{L^p(\mathbb R)}\xrightarrow{h\to0}0\tag2$$ and $$\forall h\in\mathbb R:\left\|g_{h,\:\varepsilon}-g_h\right\|_{L^p(\mathbb R)}\xrightarrow{\varepsilon\to0+}0\tag3.$$
And even if we can show $(1)$, how does the claim for $f$ follow from that? From Jose27's comment below I suppose that there must be something crucial with the endpoints ...
I'll be borrowing the notation you use in your edit, basically setting $f=\eta_1 u=v$ so that in particular $f\in E^{p,q}$ and $f$ has compact support on $[0,T)$, and so we may as well assume $I=[0,\infty)$.
We need to show that for any $\varepsilon>0$ there exists $f^\varepsilon\in C^\infty(I)\cap E^{p,q}$ such that $\| f -f^\varepsilon\|_{E^{p,q}} <\varepsilon$. The idea is to produce $h=h(\varepsilon)>0$, $\delta=\delta(\varepsilon,h)>0$ and $f_{h,\delta}\in C^{\infty}\cap E^{p,q}$ with the following properties:
$\| f- f_{h}\|_{E^{p,q}}<\varepsilon/2$, and
$\| f_h - f_{h,\delta}\|_{E^{p,q}} <\varepsilon/2$.
If we're able to produce such functions, then $f^\varepsilon:= f_{h,\delta}$ gives the desired approximation by the triangle inequality, as you mentioned in your post.
Now, the reason why we introduce the $h$ has more to do with the type of approximation that we're achieving: If we succeed, notice that $f_\varepsilon\in C^\infty(\bar{I})$, i.e. $f_\varepsilon$ is smooth up to the boundary of your interval. I'm not sure if this is intentional or not, given that they only mention $C^\infty(I; X)$, but they're proving more. To be fair though, in one dimension, it's actually easier to argue this way owing to the simple geometry of intervals; as opposed to higher dimensions, where smoothness of the boundary plays a role (See for instance the "H=W" result of Meyers-Serrin, and the failure of approximations by smooth functions up to the boundary; both topics are covered briefly in Evans's PDE book).
Now to the proof. Fix $\varepsilon>0$.
We can guarantee 1 by the Corollary in your text, by choosing $h=h(\varepsilon)$ sufficiently small.
Now let $\delta>0$ and consider $\eta_\delta$ a standard mollifier as in your post. Since $f_h: [-h,\infty)\to X$ and $\text{supp}(\eta_\delta)\subset (-\delta, \delta)$, we see that the convolution $$ f_{h,\delta}:= \eta_\delta* f_h = \int_{\mathbb{R}} \eta_\delta(y) f(x-y)\, dy, $$ is well defined for $x\geq 0$ as long as $\delta<h$, in other words, $f_{h,\delta}:[0,\infty)\to X$ and arguing as you did we can check that $f_{h,\delta}\in C^\infty(I,X)\cap E^{p,q}$. It remains to obtain the estimate in 2, for an appropriate choice of $\delta$. For this we have two terms to consider: $$ \| f_h-f_{h,\delta}\|_{L^p(I;X)}, \quad \text{and} \quad \| f_n'-f_{h,\delta}'\|_{L^q(I;Y)}. $$ The first one we can make smaller than, say, $\varepsilon/4$ by choosing $\delta$ small enough (depending on both $h$ and $\varepsilon$), owing to properties of mollifiers(+). For the second we use that convolution "commutes" with derivatives(*): $$ (\eta_\delta* f_h)' = \eta_\delta* (f_h'), $$ and so the second term can be handled exactly as the first one; possibly making $\delta$ smaller.
(+) More specifically we have: For any $1\leq p <\infty$, $a>0$, $g\in L^p([-a,\infty); X)$ and $\delta<a$, then $\eta_\delta*g$ is well defined and moreover $\| g-g_\delta\|_{L^p([0,\infty);X)} \to 0$ as $\delta\to 0$.
(*) Owing to the mismatch in spaces, let me sketch how we prove this. Recall that we're talking about weak derivatives here, so we need to show that for every $\phi\in C_c^\infty(0,\infty)$ it holds \begin{equation} \begin{split} \int_{0}^\infty \phi' \iota (\eta_\delta * f_h)\, dt & = - \int_{0}^\infty \phi \eta_\delta * f_h' \, dt \\ &= -\int_{\mathbb{R}}\int_{\mathbb{R}} \phi(t) \eta_\delta(s)f_h'(t-s)\, ds \, dt\\ & =- \int_{\mathbb{R}}\eta_\delta(s)\int_{\mathbb{R}}\phi(t)f_h'(t-s)\, dt \, ds, \end{split} \end{equation} where the use of Fubini's theorem is justified since $\phi,\eta$ are compactly supported and bounded, while $f_h'$ is in $L^p$ and has compact support. Since $\delta<h$, the last term above can be written as $$ \int_{\mathbb{R}} \eta(s) \int_{\mathbb{R}} \phi'(t) \iota f_n(t-s)\, dt \, ds, $$ owing to the definition of $f_h'$ (or more accurately the translate $f_{h-s}'$). Now it's a simple matter or reversing the order of integrals again to get the result.
As for the space $E_0^{p,q}$; it's typically defined as the closure of $C_c^\infty$ in $E^{p,q}$. Sometimes it has nice characterizations in terms of traces, and I suspect this is true in this context, but I don't have a formal proof at the moment.
Edit: This is too long for a comment so I'm putting it here: The way you extend the function doesn't work; the main issue is that you're introducing a discontinuity at $0$, since $\theta_1(0)=1$ by the requirement that $\theta_1+\theta_2=1$ (think about what would happen if you're trying to extend $f=1$ on $I=[0,1]$ for instance). What you could do is extend $f_h$ with a function $\theta_1$ whose support lies on $(-h, 1/3T)$ or something like this, but then you have to track the dependence on $h$ a bit more carefully when you regularize since derivatives will interact with your cut-off function.
In the end, it's mostly a cosmetic change though, the idea is that past $\delta=h$ you're gaining no information on $f$ at all: the regularizations start picking up values that have no relation to $f$. It's fine either way though, as long as you extend the function in the correct way.