[proofplan] The zero energy competitor is the singular map from $(0,1)$ to $\mathbb R$ given by $x\mapsto x^{1/3}$, which is absolutely continuous but has unbounded derivative at the origin. The strict gap for $C^1$ competitors follows from the stated classical one-dimensional Manià lower-bound estimate for the endpoint class $\mathcal C$. Finally, the structural properties of the integrand follow directly from continuity, nonnegativity, and convexity of the sixth power. [/proofplan] [step:Construct the singular Sobolev competitor with zero energy] Define the map $u_0:(0,1)\to\mathbb R$ by \begin{align*} u_0(x)=x^{1/3}. \end{align*} Define also the map $w:(0,1)\to\mathbb R$ by \begin{align*} w(x)=\frac{1}{3}x^{-2/3}. \end{align*} The function $w$ belongs to $L^1(0,1)$, since \begin{align*} \int_0^1 |w(x)|\,d\mathcal L^1(x)=\frac{1}{3}\int_0^1 x^{-2/3}\,d\mathcal L^1(x)=1. \end{align*} For every $x\in[0,1]$, \begin{align*} x^{1/3}=\int_0^x w(t)\,d\mathcal L^1(t). \end{align*} Hence the continuous representative of $u_0$ on $[0,1]$ is absolutely continuous, and its [weak derivative](/page/Weak%20Derivative) is represented by $w$. Thus $u_0\in W^{1,1}(0,1)$. Its continuous representative on $[0,1]$ satisfies $u_0(0)=0$ and $u_0(1)=1$, so $u_0\in\mathcal A$. For every $x\in(0,1)$, \begin{align*} u_0(x)^3-x=0. \end{align*} Therefore the energy density vanishes $\mathcal L^1$-a.e. on $(0,1)$, and hence \begin{align*} I[u_0]=\int_0^1 0\,d\mathcal L^1(x)=0. \end{align*} Since $I$ is nonnegative by definition, this proves \begin{align*} \inf_{u\in\mathcal A} I[u]=0. \end{align*} [/step] [step:Use the stated Manià lower-bound estimate] The endpoint class $\mathcal C$ and the constant $c_M$ are defined in the theorem statement, and the statement assumes the classical one-dimensional Manià lower-bound estimate $c_M>0$ for this class. Let $u\in\mathcal A\cap C^1([0,1])$. Then its Sobolev representative is continuous on $[0,1]$ and has endpoint values $u(0)=0$ and $u(1)=1$. Thus $u\in\mathcal C$. The map $x\mapsto (u(x)^3-x)^2|u'(x)|^6$ is continuous on the compact interval $[0,1]$, hence belongs to $L^1(0,1)$. Therefore the defining finite-energy branch of $I$ applies to $u$. By the definition of $c_M$, \begin{align*} I[u]=\int_0^1 (u(x)^3-x)^2|u'(x)|^6\,d\mathcal L^1(x)\ge c_M>0. \end{align*} Thus every $C^1$ admissible competitor has energy bounded below by the same positive constant. [guided] The constant $c_M$ is the infimum of the energy over the endpoint class \begin{align*} \mathcal C=\{v\in C^1([0,1]):v(0)=0\text{ and }v(1)=1\}. \end{align*} The theorem statement includes the classical one-dimensional Manià lower-bound estimate for this class, namely \begin{align*} c_M>0. \end{align*} We only need to check that the present competitor belongs to that endpoint class and that the finite integral formula for $I[u]$ applies. Since $u\in\mathcal A\cap C^1([0,1])$, the endpoint conditions in the definition of $\mathcal A$ give $u(0)=0$ and $u(1)=1$, and the additional condition $u\in C^1([0,1])$ gives $u\in\mathcal C$. Also, $x\mapsto (u(x)^3-x)^2|u'(x)|^6$ is continuous on $[0,1]$, so it is integrable with respect to $\mathcal L^1$ on $(0,1)$. Substituting $v=u$ into the variational class defining $c_M$ yields \begin{align*} I[u]=\int_0^1 (u(x)^3-x)^2|u'(x)|^6\,d\mathcal L^1(x)\ge c_M. \end{align*} Because $c_M>0$, this is a uniform positive lower bound independent of the particular $C^1$ competitor. [/guided] [/step] [step:Conclude the strict Lavrentiev gap] The previous step proves that, for every $u\in\mathcal A\cap C^1([0,1])$, \begin{align*} I[u]\ge c_M>0. \end{align*} Therefore \begin{align*} \inf_{u\in\mathcal A\cap C^1([0,1])}I[u]\ge c_M>0. \end{align*} Together with the zero-energy Sobolev competitor $u_0\in\mathcal A$, this gives \begin{align*} 0=\inf_{u\in\mathcal A}I[u]<\inf_{u\in\mathcal A\cap C^1([0,1])}I[u]. \end{align*} [/step] [step:Verify continuity, nonnegativity, and convexity of the integrand] Define the map $L:[0,1]\times\mathbb R\times\mathbb R\to[0,\infty)$ by \begin{align*} L(x,z,\xi)=(z^3-x)^2|\xi|^6. \end{align*} The maps $(x,z,\xi)\mapsto z^3-x$, $t\mapsto t^2$, and $\xi\mapsto |\xi|^6$ are continuous, and products of continuous real-valued functions are continuous. Hence $L$ is continuous. Since both factors $(z^3-x)^2$ and $|\xi|^6$ are nonnegative, $L(x,z,\xi)\ge 0$ for every $(x,z,\xi)\in[0,1]\times\mathbb R\times\mathbb R$. It remains to check convexity in $\xi$. Define the function $q:\mathbb R\to[0,\infty)$ by \begin{align*} q(\xi)=|\xi|^6=\xi^6. \end{align*} The function $q$ is $C^2$ on $\mathbb R$ and satisfies \begin{align*} q''(\xi)=30\xi^4\ge 0 \end{align*} for every $\xi\in\mathbb R$. Therefore $q'$ is nondecreasing on $\mathbb R$. If $\xi_1,\xi_2\in\mathbb R$ with $\xi_1<\xi_2$ and $\lambda\in[0,1]$, applying the [Mean Value Theorem](/theorems/631) to $q$ on $[\xi_1,(1-\lambda)\xi_1+\lambda\xi_2]$ and on $[(1-\lambda)\xi_1+\lambda\xi_2,\xi_2]$, together with the monotonicity of $q'$, gives \begin{align*} q((1-\lambda)\xi_1+\lambda\xi_2)\le (1-\lambda)q(\xi_1)+\lambda q(\xi_2). \end{align*} Thus $q$ is convex. For fixed $(x,z)\in[0,1]\times\mathbb R$, the coefficient $(z^3-x)^2$ is nonnegative, so the map \begin{align*} \xi\mapsto L(x,z,\xi)=(z^3-x)^2q(\xi) \end{align*} is a nonnegative scalar multiple of a [convex function](/page/Convex%20Function). Hence $L$ is convex in the gradient variable $\xi$. [/step]