# Two proofs of the fundamental theorem of calculus of variations - one correct, one not?

Fundamental Theorem of the Calculus of Variations. Let $$u \in L^1_{\text{loc}}(a,b)$$ and $$\int_{a}^{b} u(x) \varphi(x) dx = 0 \quad \forall \varphi \in \mathcal{C}^{\infty}_{\text{c}}(a,b).$$ Then, we have $$u|_{(a,b)} \equiv 0$$ almost everywhere.

I am going to present the proof our professor did and the one I found in the lecture notes from the same course a few years back. I would like to know if both approaches are correct or the first is not because it ignores some details or if the second proof is just unnecessarily complicated.

Proof 1: Let $$[c,d] \subset (a,b)$$ a compact interval with $$c < d$$. We aim to show $$\int_{c}^{d} | u(x) | dx = 0$$, which yields the claim since $$[c,d]$$ was chosen arbitrarily. Define $$\omega := \text{sgn}(u) \cdot \chi_{[c,d]}$$. Furthermore, define $$\omega_{\varepsilon} := \omega \star J_{\varepsilon}$$ ($$\star$$ = convolution), where $$J_{\varepsilon}(x) := \begin{cases} c_{\varepsilon}\exp\left(\frac{1}{x^2 - 1}\right), & | x | < 1, \\ 0, & \text{else.} \end{cases}$$ and $$c_{\varepsilon}$$ is chosen such that $$\int_{\mathbb{R}} J_{\varepsilon}(x) dx = 1$$.

In a previous lemma we have shown that

1. for sufficiently small $$\varepsilon > 0$$ we have $$\omega_{\varepsilon} \in \mathcal{C}^{\infty}_{\text{c}}(a,b)$$ and
2. $$\omega_{\varepsilon} \xrightarrow{\varepsilon \searrow 0} \omega$$ almost everywhere on $$(a,b)$$.

We now want to use Lebesgue's theorem to show $$\begin{equation*} \int_{a}^{b} \omega_{\varepsilon}(x) u(x) = \int_{c}^{d} | u(x) | dx = 0. \end{equation*}$$ To find a integrable majorant for $$\omega_{\varepsilon}$$ we observe that $$\begin{equation*} | \omega_{\varepsilon}(x) | = \left| \int_{\mathbb{R}} J_{\varepsilon}(x - \xi) \omega_{\varepsilon}(\xi) d\xi \right| \le \max_{\xi \in \mathbb{R}} | \omega_{\varepsilon}(\xi) | \cdot \int_{\mathbb{R}} J_{\varepsilon}(x - \xi) d \xi = 1 \cdot 1 = 1. \end{equation*}$$ Therefore, $$| u(x) \omega_{\varepsilon}(x) | \le | u(x) |$$.

Because $$u \in L^1_{\text{loc}}(a,b)$$ we have $$u \in L^1(c,d)$$ and therefore, $$| u |$$ is a integrable majorant for $$u \omega_{\varepsilon}$$. $$\square$$

Proof 2 Let $$u \in L^1_{\text{loc}}(a,b)$$ and $$[c,d] \subset (a,b)$$. Define $$\omega = \text{sgn}(u) \chi_{[c,d]}$$. Then we have $$\omega \in L^1_{\text{loc}}(a,b)$$ and $$\text{supp}(\omega) \subset [c,d]$$.

!The $$\tilde{J}_{\varepsilon}$$ is the $$J_{\varepsilon}$$ from above!

Now define $$\omega_{\varepsilon} := \tilde{J}_{\varepsilon} \ast \omega$$. Then, $$\omega_{\varepsilon} \to \omega$$ almost everywhere on $$(a,b)$$ and $$\text{supp}(\omega_{\varepsilon}) \subset [c - \varepsilon, d + \varepsilon]$$, hence $$\omega_{\varepsilon} \in \mathcal{C}^{\infty}_{\text{c}}(a,b)$$ if $$\varepsilon$$ is small enough.

We obtain \begin{align*} 0 = \int_{a}^{b} \underbrace{u(x) \omega_{\varepsilon}(x)}_{\xrightarrow{\textrm{a.e.}} u(x) \omega(x)} dx & = \int_{c - \varepsilon}^{d + \varepsilon} u(x) \omega_{\varepsilon}(x) dx \\ & = \int_{a}^{b} u(x) \chi_{[c - \varepsilon, d + \varepsilon]}(x) \omega_{\varepsilon}(x) dx. \end{align*} As $$\begin{equation*} | \omega_{\varepsilon}(x) | \le \int \tilde{J}_{\varepsilon}(x - y) \underbrace{| \omega(y) |}_{\le 1} dy \le 1, \end{equation*}$$ For $$\varepsilon_0 < \min(c - a, b - d)$$ and all $$\varepsilon < \varepsilon_0$$ we get $$\begin{equation*} | u(x) \omega_{\varepsilon}(x) | \le | u(x) | \chi_{[c - \varepsilon_0, d + \varepsilon_0]}(x) \end{equation*}$$ This function is integrable on $$(a,b)$$. Therefore, Lebesgues theorem shows $$\begin{equation*} 0 = \int_{a}^{b} u(x) \omega(x) dx = \int_{c}^{d} | u(x) | dx, \end{equation*}$$ hence $$u \equiv 0$$ almost everywhere on $$[c,d]$$. As $$[c,d] \subset (a,b)$$ was chosen arbitrarily, this yields the claim. $$\square$$

• I somehow without noticing down voted this question: If you edit it, I can up vote it. – onurcanbektas May 7 at 7:52
• @onurcanbektas done! – Viktor Glombik May 7 at 9:07

Both proofs are fine, although there is a limit missing in the first one, it should say $$\begin{equation*} \lim_{\epsilon \to 0}\int_{a}^{b} \omega_{\varepsilon}(x) u(x) = \int_{c}^{d} | u(x) | dx = 0. \end{equation*}$$ and the definition of $$\omega$$ is wrong, it should be $$\omega(x) := \text{sgn}(u(x))\chi_{[c,d]}(x).$$ The main difference between the proofs is this: the second proof contains an argument why $$\text{supp}(\omega_{\epsilon}) \subseteq (a,b)$$ for $$\epsilon$$ small enough, while in the first proof they argue this point by referring to a "previous lemma".
• Thanks for noticing the typo in the definition of $\omega$ in the first proof! – Viktor Glombik May 5 at 15:15