How can I prove that $\sup\{\Phi(f):f\in \bar B\}=1$ if $\bar B=\{f\in C\left([0,1]\right): d(f,0)\leq 1\}$?

Question

let me consider the metric $$d(f,g)=\max\left\{|f(t)-g(t)|:t\in [0,1]\right\}$$ for all $f,g, \in C\left([0,1]\right)$. Define also $$\Phi:C\left([0,1]\right)\rightarrow \Bbb{R};~~~f\mapsto |f(1)|-\int_0^1|f(t)|dt$$ Define $\bar B=\{f\in C\left([0,1]\right): d(f,0)\leq 1\}$, then I need to show that $$\sup\{\Phi(f):f\in \bar B\}=1$$

I have done the following:

1. Step: I need to show that $\Phi(f)\leq 1$ for all $f\in \bar B$. Indeed $$\Phi(f)=|f(1)|-\int_0^1|f(t)|dt\leq |f(1)|\leq 1$$ using in the first inequality that $\int_0^1 |f(t)|dt\geq 0$ and in the second that $d(f,0)\leq 1$.
2. Step: I need to show that there is no smaller upper bound. Therefore I picked $\epsilon >0$. Assuming $1-\epsilon$ is a smaller upper bound should give a contradiction. But somehow I don't get to this contradiction. I thought about chosing an explicit $f$, maybe a constant so that $|f(1)|+\epsilon<1+\int_0^1 |f(t)|dt$ gives a contradiction.

Is the first step correct and if yes how can I do the second step?

Thanks a lot.

Answer 1

Step 1 is correct. For Step 2 you can consider functions in $\bar B$ with $f(1) = 1$ and $\int_0^1 |f(t)| dt$ arbitrarily small, for example $$ f_n(x) = \begin{cases} 0 & 0 \le x \le 1-1/n \\ n x - n + 1 & 1-1/n \le x \le 1 \\ \end{cases} $$