Dirichlet's Energy on $X:=\{u\in C^1([0,1]) |\ u(0)=0, u'(1)=1\}$? Let $X:=\{u\in C^1([0,1]) |\ u(0)=0, u'(1)=1\}$. I want to show that  $J(u):= \int_0^1 (u'(x))^2dx$ doesn't have an infimum on $X$.
Hi, this looks an awful lot like an application for Dirichlet's energy. But I don't even know if $u \in C^2(0,1)$. If I assume that it is, then an infimum would be a solution of the problem $u''=0$ implying that $u'=c$ and subsequently $u'=1$. If $u'=1$, then $u(x)=x$ if I want $u(0)=0$. But where is the contradiction?
Thank you
 A: Notice that $u(x) = x$ is obviously not minimizing the energy, indeed $J(u) = 1$, but we can easily prove that $$\inf_{v \in X} J(v) = 0.$$ Indeed consider the following sequence of functions defined piecewise: 
$$
v_n(x) = 
\begin{cases}
0 & \text{if}\ x \in [0,1-\frac{1}{n}] \\
x - 1 + \frac{1}{n}  & \text{if}\ x \in [1-\frac{1}{n},1]
\end{cases}
$$
Now, $v_n$ satisfies $v_n(0) = 0$ and $v_n'(1) = 1$. Unfortunately it is not $C^1$, but the corner can be smoothened out easily. Moreover $v'_n(x) = 0$ for $x \in (0,1-\frac{1}{n})$, hence $$J(v_n) \le \int_{1-\frac{1}{n}}^1v_n'(x)^2\,dx \le C\frac 1n \to 0.$$
I'll leave it you to fill in the details in the above argument. Now that we have the value of the infimum, assume by contradiction that there is $w \in X$ for which $J(w) = \inf_X J.$ Then obviously $w' = 0$, which implies $w(x) = 0$ for all $x$ since $w(0) = 0$. This contradicts the fact that $w'(1) = 1$.
A: Suppose that $u_0\in X$ is such that $J$ has a minimum at $u_0$. Put, for $a\in \mathbb{R}$ and $\displaystyle h(x)=(1-x)^n-1$, $n\geq 2\in \mathbb{N}$, $u(x)=u_0(x)+ah(x)$. We see that $u\in X$. Then we have $J(u)\geq J(u_0)$; we get that $$2a\int_0^{1}u_0^{\prime}(x)h^{\prime}(x)dx+a^2\int_0^1 (h^{\prime}(x))^2dx\geq 0$$
Suppose first $a>0$. We divide the inequality by $a$, let $a\to 0$ and we get $$ \int_0^{1}u_0^{\prime}(x)h^{\prime}(x)dx\geq 0$$ Now suppose $a<0$, divide the inequality by $a$ and let $a\to 0$. We get $$\displaystyle \int_0^{1}u_0^{\prime}(x)h^{\prime}(x)dx\leq 0$$ hence $\displaystyle \int_0^1u_0^{\prime}(x)h^{\prime}(x)dx=0$, and we have  $\displaystyle \int_0^{1}u_0^{\prime}(x)(1-x)^{n-1}dx=0$. Now as this is true for all $n\geq 2$, and as $u_0^{\prime}(x)$ is continuous, we have $u_0^{\prime}(x)(1-x)=0$ , hence $u_0^{\prime}=0$ (on $[0,1[$ first, then on $[0,1]$ by continuity). But $u_0^{\prime}(1)=1$,  a contradiction. So there does not exists such $u_0$. 
