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

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):

enter image description here

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?

some_math_guy
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    this answers your first question – Lorago Jun 20 '23 at 14:49
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    For your second question, if you apply the "continuous" version on every segment where your $u$ is continuous, you get a constant $c_k$ per segment, and then I think you should be able to use $\mathcal{C}^\infty_c$ functions whose supports overlap two and exactly two segments to recover the fact that the $c_k$s are equal. – Bruno B Jun 20 '23 at 14:54
  • @Lorago. What is the intuition that makes my initial belief wrong? I mean, I thought that one could integrate on each continuity interval separately and chose a different integration constant on each one so as to ensure that there is no integrability – some_math_guy Jun 20 '23 at 15:55
  • @some_math_guy Note that, when integrating, we don't really care about the value of the function $u$ at some specific points (or even countably many). For a continuous function $u$ on $[a,b]$ we can always change the value at some point, say at $(a+b)/2$, to get a modified function $\tilde{u}$, which is now only piecewise continuous, but the integral of both function over $[a,b]$ is still the same. – stange Jun 20 '23 at 16:27
  • @stange : What I meant was this:. For example: let F(x)= x-3 on [-1,0] and -x+5 on [0,1] which is piecewise continuous as it is discontinuos at x=0. Let f(x)= 1 on [-1,0] and -1 on [0,1]. which I got by taking the derivative of F(x) on each interval and it's also discontinuous. Then I can regard F(x) as the integral of f(x). Why isn't this a valid example trying to show that the integral of a piecewise-continuous function is not continuous? – some_math_guy Jun 20 '23 at 16:46
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    While going from a function $f$ to its derivative $f'$, we may consider these subintervals and take the derivatives on each one of them, but we can't go from $f'$ to $f$ by just integrating on each subinterval again, compare with the fundamental theorem of analysis: To obtain the antiderivative $F$ from $f$, we have to compute $F(x) = \int_{-1}^x f(t) , \text{d}t$. As you see, the values of $f$ on $[-1,0]$ play also a role for the values of $F$ on $[0,1]$. Hope this clarifies it. – stange Jun 20 '23 at 17:16
  • @stange Yes, thank you. This clarifies that, so now is clear that $\psi(x)$is continuous. However in the proof they say even more, they say that it is in $D^1_0$. It is clear that it vanishes at the endpoints, by how do I argue that it is $C^1$ on each continuity subinterval when all I have is that $\psi$ is continuous? – some_math_guy Jun 20 '23 at 18:18
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    By the FTC, $\psi$ is differentiable with derivative $\psi' = u - c$. But as $u$ is only piecewise continuous, the same holds for $\psi'$ and hence $\psi\in D^1_0([a,b];\mathbb{R}^n)$. – stange Jun 20 '23 at 18:36
  • @stange I think you meant the same holds for $\psi'$ – some_math_guy Jun 20 '23 at 18:38
  • Yes, $\psi'$ of course. – stange Jun 20 '23 at 18:40
  • @BrunoB How would I exactly recover that two $c_k$'s are equal by using overlapping test functions? – some_math_guy Jun 20 '23 at 18:48
  • Say we name $(a_k)k$ the sequence of endpoints of the segments where $u$ is continuous. Then choose any $\mathcal{C}^\infty_c$ function $\eta_k$ such that $\eta_k(a_k) = \eta_k(a{k+2}) = 0$ and $\eta_k(a_{k+1}) = 1$, and see what happens when integrating from $a_k$ to $a_{k+2}$. – Bruno B Jun 20 '23 at 19:22
  • @BrunoB I have that $\int_{a_k}^{a_{k+1}}u\eta_1'=$, implies $u=c_k $ on that interval$\int_{a_{k+1}}^{a_{k+2}}u\eta_2'=0$ , implies $u=c_{k+1} $ on that interval where $\eta_1 \in C_c^\infty([a_k,a_{k+1}])$ and $\eta_2 \in C_c^\infty([a_{k+1},a_{k+2}])$. On the other hands for the $\eta_k $ that you describe, if I take it in $C_c^\infty([a,b])$. and as you describe it and integrate $\int_{a_{k}}^{a_{k+2}}u\eta_k'=\int_{a_{k}}^{a_{k+1}}u\eta_k' + \int_{a_{k+1}}^{a_{k+2}}u\eta_k'$, but I don't know how to relate it with the other $\eta_1$ and $\eta_2$ – some_math_guy Jun 20 '23 at 19:40
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    You don't need to relate them. I did say any function but I forgot to re-say that they should be of support contained in $[a_k, a_{k+2}]$. This way, the integral over that interval is equal to the integral over $[a,b]$, which is $0$ by assumption. – Bruno B Jun 20 '23 at 19:42

2 Answers2

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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.

stange
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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.

Bruno B
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  • I think you meant on $[a,m) \cup (m,b]$ – some_math_guy Jun 20 '23 at 20:34
  • Thanks for the heads up. – Bruno B Jun 20 '23 at 20:36
  • You're welcome, I did manage to get to this part with your suggestion, and actually I was wondering about the existence of that $\eta$. How would I prove that? Isn't it sufficient yo use the standard mollificator which is a $C_c^\infty$ function as to say this is it... or do you have another idea in mind? – some_math_guy Jun 20 '23 at 20:39
  • @some_math_guy It depends on what you'd call the standard mollificator I guess. – Bruno B Jun 20 '23 at 21:01
  • I meant mollifier. This one https://en.wikipedia.org/wiki/Mollifier#Concrete_example. One can shrink the support and change the amplitude. What did you had in mind? – some_math_guy Jun 20 '23 at 21:13
  • Right. Then yeah, you can use this mollifier, say, from $-2$ to $2$, and then choose one polynomial $p$ passing by $(a, -2)$, $(m,0)$ and $(b,2)$ so that you can transpose that specific mollifier onto the interval $[a,b]$ via $\eta = \phi \circ p$ without having to redo the proof that it's $\mathcal{C}^\infty_c$ all over again. – Bruno B Jun 20 '23 at 21:18