# The regularity theorem, a non-regular minimizer problem

During my self study to the calculus of variations I come across this problem. Because of my search, I know what I wanted to do but I need some help to do them.

The function $$f:[-1,1] \times \mathbb R \to \mathbb R$$ is defined by $$f(x,\xi) = \left[ w(x) \xi - 2 x w(x) \sin\left(\frac{\pi}{x}\right) + \pi w(x) \cos\left(\frac{\pi}{x}\right) \right]^2,$$ where $$w(x) = \begin{cases} e^{- \frac{1}{x^2}},& x\not = 0\\ 0, & x=0 \end{cases}.$$ In this problem we consider the regularity of solutions of the Euler-Lagrange equation corresponding to the one-dimensional minimization problem $$\inf_{u \in X} I[u],$$ where $$X= \big\{ u \in \operatorname{Lip}[-1,1]: u(-1) = u(1)=0 \big\}$$ and $$I[u] = \int_{-1}^1 f(x, u'(x)) dx.$$ I want to prove the so-called regularity theorem; Let $$\overline{u} \in \operatorname{Lip}[a,b]$$ satisfy the integral Euler equation for the functional $$\displaystyle \int_a^b f(x,u(x),u'(x)) dx$$, where for every $$x \in [a,b]$$ the function $$\xi \to f(x,\overline{u}(x),\xi)$$ is strictly convex. Then $$\overline{u}$$ lies in $$C^1 [a,b]$$.

As far as reached, to prove the theorem we can solve the following sub-problems, to explain why the regularity theorem does not apply to this minimization problem.

The steps are:

• Showing that $$f$$ is infinitely differentiable, $$\xi \mapsto f(x,\xi)$$ is convex and $$f_{\xi \xi} (x,\xi) > 0$$ holds for all $$x$$ except for $$x=0$$.
• Showing that the function $$\overline{u}(x) = \begin{cases} x^2 \sin \frac \pi x,& x \not = 0\\ 0, & x=0\end{cases}$$ yields the minimum of the minimization problem. Also confirm that this function is Lipschitz continuous on $$[-1,1]$$.
• And we show that there is no other minimizer except for the function $$\overline{u}$$ above.
• And we need to show that $$\overline{u}$$ does not belong to $$C^1 ([-1,1])$$.

The main step is to explain why the regularity theorem does not apply

My Proof: First we prove that the function $$w(x)$$ is infinitely differentiable. The function to be infinitely differentiable, one would need to show that in the limit the function $$w(x)$$, and all its derivatives, go to zero as $$x$$ goes to $$0$$. Since $$x \mapsto \frac{1}{x}$$ is smooth for $$x \neq 0$$ and $$x \mapsto e^x$$ is smooth, it is clear that $$w$$ is smooth for $$x \neq 0$$.

Suppose $$x \ne 0$$, then $${w}^{(k)}$$ has the form $${w}^{(k)}(x) = e^{-{1 \over x^2}} p_k({1 \over x})$$ for some polynomial $$p_k$$. This is clearly true for $$k=0$$, so suppose it is true for $$k=0,...,n$$. Then $${w}^{(n)}(x) = e^{-{1 \over x^2}} p_n({1 \over x})$$ and the chain rule gives $$\begin{split} {w}^{(n+1))}(x) & = {w}^{(1)}(x) p_n\left({1 \over x}\right) - {w}^{(0)}(x) p_n'\left({1 \over x}\right) \left({1 \over x^2}\right) \\ & = e^{-{1 \over x^2}} \left[{2 \over x^3}p_n\left({1 \over x}\right)-p_n'\left({1 \over x}\right) \left({1 \over x^2}\right) \right]. \end{split}$$ If $$p_{n+1}(y) = 2 y^3p_n(y)-p_n'(y) y^2$$, then $${w}^{(n+1)}(x) = e^{-{1 \over x^2}} p_{n+1}\left( {1 \over x} \right),$$ and so the result is true for all $$n$$. If $$x \neq 0$$, we have $$e^{-{1 \over x^2}} = {1 \over {e^{1 \over x^2}}}\quad \text{ and }\quad e^{1 \over x^2} \ge \sum_{k=0}^n {1 \over k!} {1 \over x^{2k}}$$ and thus $$e^{-{1 \over x^2}} \le {x^{2n} \over \sum_{k=0}^n {1 \over k!} {x^{2(n-k)}}} \le {x^{2n} \over n!}.$$ Suppose $$p$$ is a polynomial of degree $$d$$. Then for any $$n$$ we see that there is some constant $$K$$ such that $$|e^{-{1 \over x^2}} p({1\over x})| \le K |x|^{2n-d}$$ whenever $$0 <|x| \le 1$$. In particular, there is some $$K$$ such that $$|e^{-{1 \over x^2}} p({1\over x})| \le K x^2$$ for all $$0 < |x| \le 1$$. We have $${w}^{(0)}(x) \le x^2$$ for all $$x$$, and so $${w}$$ is continuous at $$x=0$$. Since $$|{w}^{(0)}(x) - {w}^{(0)}(0) -0| \le x^2$$, we see that $${w}^{(0)}$$ is differentiable at $$x=0$$, and $${w}^{(1)}(0) = 0$$.

Now suppose $${w}^{(k)}$$ is differentiable at $$x=0$$ and $${w}^{(k)}(0) = 0$$ for $$k=0,...,n$$. Then $$|{w}^{(n)}(x) - {w}^{(n)}(0) -0| \le K x^2$$ for some $$K$$ and $$|x| \le 1$$. Hence $${w}^{(n)}$$ is differentiable at $$x=0$$, and $${w}^{(n+1)}(0) = 0$$. With a similar argument we find that the function

$$\cos {\pi \over x} + i \sin \frac \pi x= \begin{cases} e^{- \frac{\pi}{x}},& x\not = 0\\ 0, & x=0 \end{cases}$$ is infinitely differentiable. Therefore, $$\cos{\frac \pi x}$$ and $$\sin \frac{\pi}{x}$$ are smooth functions. Since the summation and product of smooth functions is smooth function then $$f$$, defined above, is smooth.

Next step is showing $$f(x,\xi)$$ is convex w.r.t. $$\xi$$. For notational simplicity we put $$g=2 x \sin \frac{\pi}{x} + \pi \cos\frac{\pi}{x}.,$$ then for fix $$x \in [-1,1]$$, we have $$f(\xi)= w^2(\xi-g)^2$$ is a convex function, for let $$\lambda\in [0,1]$$, then $$\begin{split} f&(x,\lambda \xi_1 + (1-\lambda) \xi_2) - \lambda \xi_1 f( \xi_1) - (1-\lambda) f(\xi_2) \\ & = w^2(\lambda \xi_1 + (1-\lambda) \xi_2 -g)^2 - \lambda w^2( \xi_1-g)^2 - (1-\lambda) w^2(\xi_2-g)^2 \\ & = w^2(\lambda (\xi_1-g) + (1-\lambda) (\xi_2 -g))^2 - \lambda w^2( \xi_1-g)^2 - (1-\lambda) w^2(\xi_2-g)^2 \\ &=w^2(\lambda^2 (\xi_1-g)^2 +2\lambda(1-\lambda)(\xi_1-g)(\xi_2 -g) + (1-\lambda)^2 (\xi_2 -g)^2 -\lambda( \xi_1-g)^2 -(1-\lambda)(\xi_2-g)^2 )\\ & = w^2( (\lambda^2-\lambda)(\xi_1-g)^2 + 2\lambda(1-\lambda)(\xi_1-g)(\xi_2 -g) + ((1-\lambda)^2 - (1-\lambda) ) (\xi_2 -g)^2)\\ & = w^2(\lambda(1-\lambda)(\xi_1-g)^2 +2\lambda(1-\lambda)(\xi_1-g)(\xi_2 -g) + \lambda(1-\lambda) (\xi_2 -g)^2)\\ & = \lambda(1-\lambda)w^2( (\xi_1-g)^2 + 2)(\xi_1-g)(\xi_2 -g) + (\xi_2 -g)^2) \\ &= \lambda(1-\lambda)w^2((\xi_1-g) + (\xi_2 -g))^2 \geq 0 \end{split}$$ (The last inequality is as a product of two non-negative numbers and two square numbers is non-negative)

$$\newcommand\ol\overline$$The steps were:

1. Show that $$f$$ is infinitely differentiable, $$\xi \mapsto f(x,\xi)$$ is convex and $$f_{\xi \xi} (x,\xi) > 0$$ holds for all $$x$$ except for $$x=0$$.

2. Show that the function $$$$\overline{u}(x) = \begin{cases} x^2 \sin \frac \pi x,& x \not = 0\\ 0, & x=0\end{cases} \tag{0}$$$$ yields the minimum of the minimization problem. Also confirm that this function is Lipschitz continuous on $$[-1,1]$$.

3. Show that there is no other minimizer except for the function $$\overline{u}$$ above.

4. Show that $$\overline{u}$$ does not belong to $$C^1 ([-1,1])$$.

Step 1: You have already checked that $$f$$ is infinitely differentiable. Next, $$\begin{equation*} f(x,\xi)=w(x)^2(\xi - g(x))^2, \tag{1} \end{equation*}$$ where $$\begin{equation*} g(x):=2 x \sin\frac{\pi}{x} - \pi \cos\frac{\pi}{x}. \end{equation*}$$ Note that $$f(x,\xi)$$ is so far undefined at $$x=0$$ (since $$g(x)$$ is so far undefined at $$x=0$$). So, let $$g(0):=0$$ and $$f(0,\xi):=0$$, so that (1) holds even for $$x=0$$. Then clearly $$\xi \mapsto f(x,\xi)$$ is convex and $$f_{\xi \xi} (x,\xi)=2w(x)^2 > 0$$ for all $$x\ne0$$.

Steps 2 and 3: We have $$I[\ol u]=0$$, since $$\begin{equation*} I[u] = \int_{-1}^1 w(x)^2(u'(x) - g(x))^2\, dx \end{equation*}$$ and $$\ol u'=g$$. Also, if $$u\in X\setminus\{\ol u\}$$, then the Lebesgue measure of the set $$\{x\colon u'(x)\ne g(x)\}=\{x\colon u'(x)\ne\ol u'(x)\}$$ is $$>0$$ and hence $$I[u]>0$$. So, $$\ol u$$ is the only minimizer of $$I[u]$$.

Also, $$|\ol u'(x)|=|g(x)|\le2|x|+\pi\le2+\pi$$ for $$x\in[-1,1]\setminus\{0\}$$ and $$\ol u$$ is continuous. So, $$\ol u$$ is Lipschitz continuous on $$[-1,1]$$.

Step 4: By the definition of the derivative and (0), $$\ol u'(0)=0$$. However, $$\ol u'(x)$$ does not converge as $$x\to0$$. Indeed, otherwise, $$-\pi \cos\frac{\pi}{x}=g(x)-2 x \sin\frac{\pi}{x}=\ol u'(x)-2 x \sin\frac{\pi}{x}$$ would converge as $$x\to0$$, which is clearly not so. So, $$\ol u\notin C^1 ([-1,1])$$.

This completes the steps.

• Thank you very much for your help. The main question, which is why the regularity theorem does not apply, here is still not answered explicitly. Is is because the function $f$ is not strictly convex? Or there is another reason? Jan 23 at 2:43
• @SelfLearner : I thought your question was "Could you please help with the other steps!". Anyhow, $f(x,\xi)$ is indeed not strictly convex in $\xi$ is $x=0$. Jan 23 at 3:14
• Thank you very much. I just wanted to make sure I am right. I think in your comment you meant if (not is )$x=0$. Jan 23 at 3:26
• @SelfLearner : Indeed. Sorry for the typo in the comment. Jan 23 at 3:29