Another minimization problem from my calculus of variation class. This is the last point of a longer exercise. Let's fix some notations $X=\{u\in C^{1}([0,1]),\ \ u(0)=u(1)=0\}$, $\ F:X\rightarrow\mathbb{R}$, $F(u)=\int_{0}^{1} \vert u'\vert+3\vert u-1\vert dx$. I have to show that $F$ has no minimum in $X$.
Thoughts
- $F\geq0$. If $v$ is a function on $[0,1]$ which is 1 a.e. then $F(v)=0$, thus we have a minimum for $F$ which is not in $X$.
- The regularity of the associated Lagrangian makes meaningless to speak about the Euler-Lagrange equations.
- At the very beginning I thought that if I had been able to construct a sequence of functions in $X$ converging to something like $u(x)=1$ on $(0,1)$, $u(0)=u(1)=0$, then the functional on this sequence would have converged to 0. I tried with 3 sequences with the above property, say $u_h$ one of them, but for all my choices I obtained $\lim_{h\rightarrow} F(u_h)=2$, thus I only concluded that $inf_X F\leq2$. For curiosity I then tried to compute the value of $F$ on other elements of X and I always found something larger or equal to two.
- I assumed the existence of a minimum and then I tried to get a contradiction but I got stuck almost immediately since the only information I have is $\inf_X F\leq2$.
- I noticed that the real aspect of $F$ is $F(u)=\vert\vert u'\vert\vert_{L^1}+3\vert\vert u-1\vert\vert_{L^1}$ but I'm not sure if any integral inequality can help.
In conclusion I have no idea how to tackle this part of the problem.
Since this is homework I'm looking for a hint or an idea.
Let $a=\max u=u(c)$ and say that $0\le a\le 1$. Then $$\int_0^c|u^\prime|dx\ge \left\vert\int_0^c u^\prime dx\right\vert=a$$ and $$\int_c^1|u^\prime|dx\ge \left\vert\int_c^1 u^\prime dx\right\vert=a$$ and so $$ \int_0^1|u^\prime|dx\ge 2a.$$ On the other hand, $$\int_0^13|u-1|dx \ge 3(1-a).$$ Hence $$F(u)\ge 2a+3(1-a)=3-a\ge 2.$$ So $\inf F\ge 2$
If $a>1$ you can truncate (this has to be made precise, if you are allowed to work with Lipschitz this is fine, otherwise you have to use mollifiers) $u$ and if $a<1$ then you pay at least $3$.