# Finding an infimum and showing it is never obtained for a function on a normed vector space.

I am struggling with the last part of a question about normed vector spaces.

This is the whole question:

a) Show that the set $A\subset C[0,1]$ defined by $A=\left\{f:0\leq f(x)\leq 1 \forall x\in [0,1]\right\}$ is closed in the norm $\left \| \cdot \right \|_\infty$.

b) Show that $F:A\to\mathbb{R}$ defined by $F(f)= \int^\frac{1}{2}_0f+\int^1_\frac{1}{2}(1-f)$ is a continuous function.

c) Show that $inf_{f\in A}F(f)=0$, but that this infimum is never attained on A.

This is what I have done for part a) and b) but I am struggling with c)

a) Let $\left\{f_n\right\}$ be a convergent sequence in A converging to some $f_0\in C[0,1]$. To show that A is closed need to prove that $f_0\in A$. So we know that as $\left\{f_n\right\}\in A\space\space\forall n\in\mathbb{N}$ $$0\leq f_n\leq1 \implies 0\leq \left \| f_n \right \|_\infty\leq1$$ Then by taking limits and using the fact that the sup norm is a continuous norm, it follows that:$$0\leq \left \| \lim_{n\to\infty} f_n \right \|_\infty\leq1\implies 0\leq \left \| f_0 \right \|_\infty\leq1$$ hence $$0\leq f_0\leq1\implies f_0\in A$$ so A is closed over the sup norm.

b) For this I simply split up the. integral and said that as $f\in C[0,1]$ by the FTC and the summation rules for continuous functions it follows that the function F is also continuous.

Then for part c) I was trying to do something with open and close sets but I am not getting any where and am not really sure now if I have done parts a) and b) correctly. Any hints, clarifications or solutions would be helpful! Thank you in advance!

For part c), if some $f_{0}$ is such that $F(f_{0})=\inf F(f)=0$, as $f_{0},1-f_{0}\geq 0$, then both $\displaystyle\int_{0}^{1/2}f_{0}$ and $\displaystyle \int_{1/2}^{1}(1-f_{0})$ equal to zero. As $f_{0}$ and $1-f_{0}$ are continuous, then $f_{0}(x)=0$ for all $x\in[0,1/2]$, $1-f_{0}(x)=0$ for all $x\in[1/2,1]$, so $f_{0}(1/2)=0$ and $f_{0}(1/2)=1$ at the same time.
Just consider $\{f_n\}$, where $f_n$ is defined by
$$f_n= \begin{cases} 0 & 0\leq x \leq 1/2 -1/n\\ \dfrac{1}{4}(2 n x - n + 2) & 1/2-1/n \leq x \leq 1/2+1/n\\ 1 & 1/2+1/n \leq x \leq 1 \end{cases}$$ Obviously, $f_n \in A$ $\forall n \in \mathbb{N}$ and $F(f_n) = \frac{1}{2n}\to 0$ when $n\to \infty$. Using this and $F(f)\geq 0$, $\inf_{f\in A} F(f) =0$. To prove it is not minimum, you can use the argument of @user284331.