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Let $I= (0, 1)$. Let $u \in L^2 (I)$ such that $u'' \in L^2 (I)$. Below is a proof (from this answer) that $u' \in L^2 (I)$.

Let $v (x) := \int_0^x u'' (s) \, \mathrm d s$ for $x \in I$. By Hölder's inequality, $|v(x)| \le \| u'' \|_{L^2} \sqrt x$. Then $v \in L^2 (I)$. For $\varphi \in C^\infty_c (I)$ we have $$ \begin{align*} \int_I v \varphi' &= \int_I \left ( \int_0^x u'' (s) \, \mathrm d s \right ) \varphi' (x) \, \mathrm d x \\ &= \int_I \left ( \int_s^1 \varphi' (x) \, \mathrm d x \right ) u'' (s) \, \mathrm d s \quad\text{by Fubini's theorem}\\ &= -\int_I \varphi (s) u'' (s) \, \mathrm d s \quad\text{by FTC}\\ &= -\int_I \varphi'' (s) u (s) \, \mathrm d s, \end{align*} $$ which implies $$ \int_I u \varphi' = -\int_I v \varphi \, \mathrm d s. $$

It follows that $v = u'$.

My question It follows from the construction of $v$ that $v(0)=0$ (for every such $u$). This does not make sense for me.

Could you elaborate on my confusion?

Thank you so much for your help!

Akira
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  • In the third line of your integration, you applied the FTC on the integral between $s$ and $1$. We should evaluate this as $\varphi(1) - \varphi(s)$, however this brings up a bigger question. It is my understanding that the corollary to the FTC requires our function to be defined on a closed interval, however the function $\varphi$ was only defined on the open interval $I$. – Jensense Nov 05 '23 at 18:40
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    @Jensense the support of $\varphi$ is compact and contained in $I$, so we can extend it to the whole $\mathbb R$. – Akira Nov 05 '23 at 19:53
  • Ahh, thank you! I had forgotten that tidbit; it seems to me that this still depends upon the extension at 1, however. – Jensense Nov 05 '23 at 19:57

1 Answers1

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The confusion comes from when you conclude: "which implies...". This is not true, what you can say is $$ \int_I v\varphi' \, dx = -\int_I u\varphi''\, dx,\qquad \varphi\in C_c^\infty. $$ Notice that if $v$ satisfies the above, then so does $v+C$ for any constant $C$. This is not true of your conclusion: Only one choice of constant $C$ should give an actual derivative of $u$!

Fix $\varphi\in C_c^\infty(I)$. We want to show that for some $C\in \mathbb{R}$ (independent of $\varphi$ of course) we have $$ \int_I (v-C)\varphi\, dx =-\int_I u\varphi'\, dx. $$

Fix $\eta\in C_c^\infty(I)$ with $\int_I \eta=1$ and set $\phi:=\varphi-\varphi_I\eta$, which is in $C_c^\infty(I)$ and has zero average so that $\psi(x):=\int_0^x \phi(t)\, dt$ is also in $C_c^\infty(I)$ and so using $\psi$ as our test function we get $$ \int_Iv(\varphi-\varphi_I\eta)\, dx = -\int_I u(\varphi'-\varphi_I\eta')\, dx. $$ Rearranging terms we see $$ \int_I \left(v-\int_I(v\eta+u\eta')\, dy\right)\varphi\, dx = -\int_I u\varphi'\, dx. $$

Jose27
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