1
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

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 $$ f_\delta(x)=\begin{cases} \dfrac{2}{\delta}\,x-1 & 0\le x\le\delta,\\ 1 & \delta<x\le1. \end{cases} $$

$\endgroup$
1
$\begingroup$

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 $lim_{n \rightarrow \infty} \mathcal I(g_n) = 3$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.