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Let $a, b \in \mathbb R^n$ and $f, g \in L^1 [0,1]$. Assume for all $h \in AC[0,1 ]$ (space of absolutely continuous functions) following integral equality holds

$$ \int_{0}^{1} \langle f(t) , h(t) \rangle \; dt + \int_{0}^{1} \langle g(t) , h' (t) \rangle \; dt + \langle h(0) , a \rangle + \langle h(1) , b \rangle = 0 $$

My question : Can we simplify the above expression more? In the sense above can be equivalently written in the form of an ODE in terms of $f,g,a,b$?

Note that all functions are from $[0,1]$ to $\mathbb R^n$. However we can assume they are just real valued function for simplicity. I was thinking about trying $h = \text{constant or exponential functions.}$

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    $\begingroup$ This questions seems a bit too elementary for MO; MSE would have been a better match. The basic idea is that you can integrate by parts in the second integral and then use that if $\int uh = 0$ for all $h$, then $u=0$ (it's a bit more complicated than that because of the boundary terms, but these can be handled similarly, by noting that $h(0)$, $h(1)$ are almost independent of what $h$ does otherwise). $\endgroup$
    – Christian Remling
    Sep 12, 2019 at 20:20
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    $\begingroup$ @ChristianRemling I already tried MSE but couldn't get suitable answer. $f,g$ lies in $L^1$ how can we use the integration by part? $\endgroup$
    – Red shoes
    Sep 12, 2019 at 21:08
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    $\begingroup$ crossposted: math.stackexchange.com/q/3354286/195021 $\endgroup$
    – Konstantinos Kanakoglou
    Sep 12, 2019 at 21:20
  • $\begingroup$ I think if $g$ is not essentially bounded, then $g h'$ need not be integrable. $\endgroup$
    – Mateusz Kwaśnicki
    Sep 12, 2019 at 21:21

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This is elementary but let's state it. First one only considers smooth test functions $h$ with compact support (so that the last two terms disappear). If $F(x):=\int_0^xf(t)dt$, integrating by parts the first term gives $\int_0^1\langle (F-g), h'\rangle dt=0$ for all these $h$, whence $F-g$ is a constant, that is, $g$ is (has a representant which is) $AC$ and $f$ is its weak derivative. The initial relation then writes $\langle h(1),(b+g(1))\rangle +\langle h(0),( a-g(0))\rangle=0$ for all $h\in AC$, that is, $g(0)=a$, $g(1)=-b$. In conclusion, the initial weak relation (for a continuous representant of $g$) is equivalent to: $g$ is $AC$, $f=g'$, $g(0)=a$, $g(1)=-b$.

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  • $\begingroup$ Thank You . Just one question for the part " $F-g$ is a constant, that is, $g$ is (has a representant which is) $AC$ and $f$ is its weak derivative" are you using a lemma or something? $\endgroup$
    – Red shoes
    Sep 13, 2019 at 20:12
  • $\begingroup$ Also In final condition $f =g'$, the $g$ is used might not be the original but rather a $AC$ representative of it. Is there any explicit formula that can write this AC -$g$ in terms of our orginal $g$ ? Thanks for time $\endgroup$
    – Red shoes
    Sep 13, 2019 at 20:28
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    $\begingroup$ Yes, if an element of $L^1_{loc}$ (or even a distribution) has distributional derivative zero, then it is a constant (or constant a.e., if you see it as a function). The other point: for instance a canonical AC representative of $g$ is $$\tilde g(x):=\int_0^xg'(t)dt+\int_0^1 \big(g +(t-1)g' \big) dt.$$ It is $AC$, its derivative is equal to $g'$ a.e, so $\tilde g$ and $g$ differs by a constant a.e. but the constant is 0 because $\tilde g$ and $g$ have the same integral on $ [0,1]$ $\endgroup$
    – Pietro Majer
    Sep 14, 2019 at 0:15
  • $\begingroup$ Thanks. In the formula you wrote how do we know $g'$ exists almost everywhere ? If it does exist then I think we can write the precise conclusion in the form $$ g(0)=a , \quad g(1)= -b, \quad f(t) = g'(t) \quad \quad a.e ~ t \in [0,T] $$ Where I used the same original $g$ in question . Am I right? $\endgroup$
    – Red shoes
    Sep 14, 2019 at 0:52
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    $\begingroup$ Correct. Also note that your hypotheses are invariant for a.e. perturbations, so for a pair f,g just satisfying your hypotheses, without other assumptions, we can say nothing on the point-wise side. $\endgroup$
    – Pietro Majer
    Sep 15, 2019 at 7:54

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