Is the following proof of the Du Bois-Reymond lemma valid?

Question

I am suspicious about the validity of the following proof. I have two doubts that are bugging me:

1) If the proof, is it true that $\psi \in D_0^1([a,b];\mathbb{R}^n)$? I mean $\psi$ is the integral of $u$ : $\psi = \int_a^x (u(s)-c)ds$ which is is piecewise-continuous, so in the eventual discontinuity points, also the integral is discontinuous, isn't it? So why would the $\psi$ be necessarily continuous?

Note : $D_0^1([a,b];\mathbb{R}^n)$ means functions in $D^1([a,b];\mathbb{R}^n)$ that vanish the endpoints. And $D^1([a,b];\mathbb{R}^n)$ is the class of piecewise continuously differentiable functions (which are continuous by definition):

2) Is the statement actually correct? I mean,everywhere I look I can just find a version of the lemma for $u$ continuous and not a single one for $u$ piecewise continuous as in my case. Where exactly in the proof is that hypothesis used? (I guess only in the step regarding my first question only?) Do you know where can I find the proof for my case?(In case this one is correct, it is missing the first step that they argue follows by an approximation argument) Or can you provide one?

Answer 1

As the link provided by @Lorago answers your first question, I only comment on your second one.
Yes, the statement is true. Here, the hypothesis makes sure that $u$ is actually integrable and simplifies the proof. Using more machinery, we can proof the statement even with less regularity:

$\textbf{Theorem:}$ Suppose that $f\in L^1_{\text{loc}}(a,b)$ for some $a<b\in\mathbb{R}$ and $$ \int_a^b f\eta' \, \text{d}x = 0$$ for all $\eta\in C_c^\infty(a,b)$. Then $f\equiv \text{const.}$ almost everywhere.

A proof can be found in "One-dimensional Variational Problems: An Introduction" by Buttazzo, Giaquinta and Hildebrandt, Lemma 1.8. Alternatively you can find the proof here https://mccuan.math.gatech.edu/courses/7581/notes/lecture3.pdf, lemma 4.

Answer 2

Let me make what I said in the comments into a proper answer:
Assume known the same theorem but with $u$ being continuous instead of just piecewise continuous, and let's show how it implies the piecewise continuous version.

Let $u$ be piecewise continuous satisfying the hypothesis. WLOG, we can assume that $u$ is continuous on $[a,m]$ and $[m,b]$ for some $m \in (a,b)$ (the general case will follow by looking at every pair of consecutive segments).
By the "continuous Du Bois-Raymond" on $[a,m]$ and $[m,b]$, which we can apply because $\mathcal{C}_c^\infty([a,b])$ contains $\mathcal{C}_c^\infty([a,m])$ and $\mathcal{C}_c^\infty([m,b])$, there exist two constants $c_1$ and $c_2$ such that $u \equiv c_1$ on $[a, m)$ and $u \equiv c_2$ on $(m,b]$. Note that we can't say anything at $m$: we've considered two different restrictions of $u$ to which we applied the theorem, $u|_{[a,m]}$ and $u|_{[m,b]}$ are equal to $c_1$ and $c_2$ on closed intervals but there would be no contradiction in having two constants for the starting $u$ at this step.

Now consider $\eta$ a $\mathcal{C}^\infty_c(a,b)$ such that $\eta(m) \neq 0$ (the existence of $\eta$ is left as an exercise).
Then, we get: $$\begin{split}0 = \int_a^b u(x)\eta'(x)\mathrm{d}x &= \int_a^m c_1\eta'(x)\mathrm{d}x + \int_m^b c_2\eta'(x)\mathrm{d}x\\ &= c_1(\eta(m) - \eta(a)) + c_2(\eta(b) - \eta(m))\\ &= (c_1 - c_2)\eta(m)\end{split}$$

Hence $c_1 = c_2 =: c$, and $u = c$ on $[a,m) \cup (m,b]$. Using the definition of piecewise-continuity at $m$, $u$ is thus equal to $c$ on $[a,b]$.
Thus, if the theorem is proven for continuous functions $u$, you can extend it to piecewise continuous functions.