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?

  • 3
    $\begingroup$ this answers your first question $\endgroup$
    – Lorago
    Jun 20, 2023 at 14:49
  • 2
    $\begingroup$ 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. $\endgroup$
    – Bruno B
    Jun 20, 2023 at 14:54
  • 1
    $\begingroup$ 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. $\endgroup$
    – stange
    Jun 20, 2023 at 17:16
  • 1
    $\begingroup$ 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 $\psi\in D^1_0([a,b];\mathbb{R}^n)$. $\endgroup$
    – stange
    Jun 20, 2023 at 18:36
  • 1
    $\begingroup$ 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. $\endgroup$
    – Bruno B
    Jun 20, 2023 at 19:42

2 Answers 2


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.


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.

  • $\begingroup$ I think you meant on $[a,m) \cup (m,b]$ $\endgroup$ Jun 20, 2023 at 20:34
  • $\begingroup$ Thanks for the heads up. $\endgroup$
    – Bruno B
    Jun 20, 2023 at 20:36
  • $\begingroup$ 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? $\endgroup$ Jun 20, 2023 at 20:39
  • $\begingroup$ @some_math_guy It depends on what you'd call the standard mollificator I guess. $\endgroup$
    – Bruno B
    Jun 20, 2023 at 21:01
  • $\begingroup$ I meant mollifier. This one en.wikipedia.org/wiki/Mollifier#Concrete_example. One can shrink the support and change the amplitude. What did you had in mind? $\endgroup$ Jun 20, 2023 at 21:13

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .