Show that C is a closed convex subset and its element of minimum norm

Question

I have a lot of problems with the following exercise that I can't solve.

Let $(L^1((0,1)), \|\cdot\|_{L^1}):=(E,\|\cdot\|)$ and $$C:=\{u\in E:u\geq0 \text{ a.e } x\in (0,1),\quad T(u)\geq 1\},$$ where $T:E\rightarrow \Bbb R, u\mapsto T(u)=\int_0^1xu(x)dx$. Show that:

• $C$ is nonempty, closed, convex subset of $E$;
• $d(0,C)=\inf\{\|u\|:u\in C\}=1$;
• there is no $u\in C$ such that $\|u\|=1$.

$C$ is nonempty because $u(x):=2\in C$ and also it is easy to see that $C$ is convex. My first problem is to show that $C$ is closed. So, we take $$(u_h)_h\subset C: u_h\rightarrow u \text{ in } E$$for some $u\in E$ and we want to show that $u\in C$. Now, I don't know if the following arguments are correct; I'd like if someone could help me to fix them:

1. $u\geq 0\text{ a.e. } x\in(0,1)$ because it is the limit of a.e. non negative function (??);
2. $T$ is continuous, so since $T(u_h)\rightarrow T(u)$ and $T(u_h)\geq 1$ for each $h$, it follows that also $T(u)\geq 1$.

Now let $u\in C$: $1\leq \int_0^1 xu(x)dx\leq \int_0^1u(x)dx=\|u\|$, so $1\leq d(0,C)$. To prove the equality, I guess we should find out a sequence $(u_h)_h\subset C$, such that $\|u_h\|\rightarrow 1$, but I don't know how to do it.

I have really no idea of how to approach the third and last point.

Can someone help me, please?

Answer 1

The idea to show that $d(0,C)=1$ holds, is to use a sequence of suitably chosen characteristic functions. Define the sequence $u_n$ by $$ u_n = \frac{2n^2}{2n-1}\chi_{[1-1/n,1]}(x), $$ where $\chi_{[1-1/n,1]}$, is the characteristic function of the interval $[[1-1/n,1]$. Then $u_n \ge0$. $$ T(u_n) = \frac{2n^2}{2n-1}\int_{1-1/n}^1 x dx = \frac{2n^2}{2n-1}\frac12 (1- (1-1/n)^2) = 1. $$ And $$ \|u_n\|_1 = \frac{2n^2}{2n-1}\frac1n\to 1. $$ This proves $d(0,C)\le1$.

Now assume there is $u\in C$ with $\|u\|_1=1$. Then with $\delta\in(0,1)$ $$ \begin{split} 1&\le T(u) = \int_0^1xu(x)dx = \int_0^\delta xu(x)dx + \int_\delta^1 x u(x)dx\\ &\le \delta \int_0^\delta u(x)dx + \int_\delta^1 u(x)dx\\ &= \int_0^1 u(x)dx + (\delta-1) \int_0^\delta u(x)dx \le \int_0^1 u(x)dx =1. \end{split} $$ Hence $\int_0^\delta u(x)dx=0$ for all $\delta\in(0,1)$. This together with $u\ge0$ implies $u=0$, which is a contradiction.

Both statements together imply $d(0,C)=1$. And the infimum is not attained.