supremum of a function with integral Let $\mathcal F$ be the set of continuous functions $f:[0,1]\to \mathbb R$ and $max_{0\le x\le1} |f(x)|=1$. Now let $\mathcal I:\mathcal F\to R,\mathcal I(f)=\int_0^1f(x)dx-f(0)+f(1)$. Firstly I had to show $\mathcal I(f)<3,\forall f \in \mathcal F$ and this is pretty easy. Now I have to determine $\sup({\mathcal I(f)}|f\in \mathcal F)$. Doesn't result that the supremum is $3$ from the statement before? I know that I am surely wrong. Can somebody give me some tips on how can I get the supremum, please? 
 A: The fact that $\mathcal I(f)<3$ implies that $\sup\{\mathcal I(f)\mid f\in \mathcal F\}\le3$. To show that the supremum is equal to $3$, you have to find for every $\epsilon>0$ a function $f\in\mathcal F$ such that $\mathcal I(f)\ge3-\epsilon$. This can be done for instance with functions like
$$
f_\delta(x)=\begin{cases}
\dfrac{2}{\delta}\,x-1 & 0\le x\le\delta,\\
1 & \delta<x\le1.
\end{cases}
$$
A: obvious part is that $\forall f\in \mathcal F $
$$|\mathcal I(f) | \leq |\int_0^1f(x)dx|+|f(0)|+|f(1)| \leq 1+1+1\leq 3$$
inequality here is not strict ( $\leq$ instead of $<$)
Thus, we have shown that $\sup({\mathcal I(f)}|f\in \mathcal F) \leq 3$
To prove that it is equal to 3 we need to find a function $g$ or a sequence $g_n$ of functions in $\mathcal F $, such that 
$$\mathcal I(g) = 3; \quad or \quad lim_{n \rightarrow \infty} \mathcal I(g_n) = 3$$
(this would prove that $\sup({\mathcal I(f)}|f\in \mathcal F) \geq 3$, which would solve the problem)
First thing that comes to mind is the function $g$ defined as $g(t) = 1, \; \forall t \in (0, 1]$ and $g(0)=-1$. Then $\mathcal I(g)$ would be equal to 3, but function $g$ is not contionious.
Only one step is left to complete the solution: approximate $g$ with a sequence $g_n$ of functions from $\mathcal F$ such that $lim_{n \rightarrow \infty} \mathcal I(g_n) = 3$
