Comparing $L^\infty$ norm of $f$ from $0$ to $s$ to $f(s)$.

Question

Given a function in $L^\infty(0,1)$. Is it possible to have $$\int_0^1 \|f\|_{L^\infty(0,s)}ds < \int_0^1 f(s)\, ds\ ?$$

EDIT: I do not think this is a trivial question. We know that $f(t)\leq \|f(s)\|_{L^\infty(0,s)}$ for almost every $t \in (0,s)$ based on the definition of the $L^\infty$ norm, but this does not mean that for almost every $t \in (0,s)$, $f(t) \leq \|f(t)\|_{L^\infty(0,t)}$ because $\|f(t)\|_{L^\infty(0,t)}$ could be smaller than $\|f(s)\|_{L^\infty(0,s)}$.

I am trying to prove the following: If $f \in L^\infty(0,1)$ then $$\int_0^1 \|f\|_{L^\infty(0,s)}ds \geq \int_0^1 f(s) ds.$$

I tried to show that this is true using a proof by contradiction. After assuming $$\int_0^1 \|f\|_{L^\infty(0,s)}ds < \int_0^1 f(s) ds,$$

I said that this implies there exists a set S of positive measure such that m(S)>0 and for all $s \in S$, $\|f\|_{L^\infty(0,s)} < f(s)$. This does imply that $f$ is not continuous as $s$, but I can't find a contradiction.

I also tried to show this directly. I defined simple functions $s_n=\sum_{k=1}^n a_k 1_{\Omega_k}+ \sum_{k=1}^n b_k 1_{\hat{\Omega_k}}$ where $x_k=k/n$, $$\hat{\Omega_k}=\{s \in [x_{k-1}, x_k]:f(s) >\|f\|_{L^\infty(0,x_k)}\},$$

$$\Omega_k=[x_{k-1},x_k] \setminus \hat{\Omega_k},$$

$b_k=\sup_{s \in \hat{\Omega_k}} f(s)$, and $a_k=\|f\|_{L^\infty(0,x_k)}$.

I wanted to use the fact that $s_n$ are simple functions greater than $f$, so $\int_0^1f dx<\int_0^1 s_n dx$ for all $n \in \mathbb{N}$. Then I was going to compare the limit of $s_n$ to $$\|f\|_{L^\infty(0,s)}.$$

However, I then noticed that if $f$ is a piecewise constant function that is 0 on $[0,1/2]$ and $1$ on $(1/2,1]$ then $\lim_{n \in \infty} s_n(1/2) = 1$, but $\|f\|_{L^\infty(0,1/2)}=0$.

Any other ideas of properties of $L^\infty$ functions that I can use to show this?

Answer 1

Let $f\in L^\infty(0,1)$, then $f(s) \leq \Vert f \Vert_{L^\infty(0,s)}$ for a.e. $s\in (0,1)$ and hence $$ \int_0^1 f(s) \, \text{d}s \leq \int_0^1 \Vert f \Vert_{L^\infty(0,s)} \,\text{d}s.$$

Answer 2

Following from @stange's answer, all we have to be show is that $f(s)\leq||f||_{L^{\infty}_{\left[0,s\right]}}$ a.e.

For each $n$ partition $\left[0,1\right]$ into dyadic intervals $\{\left[a^{n}_{k},b^{n}_{k}\right]\}_{k}$. Define $$ g_{n}(s) = \begin{cases} ||f||_{L^{\infty}_{\left[0,a_{k+1}^{n}\right]}} & s \in \left[a^{n}_{k},a_{k+1}^{n}\right) \end{cases} $$ We know that for each $n$, $f(s)\leq g_{n}(s)$ a.e.

Because $g(s)=||f||_{L^{\infty}_{\left[0,s\right]}}$ is non-decreasing, its easy to see $g_{n}\to g$ a.e. , it follows that $f(s)\leq g(s)$ a.e.

I do agree this isn't trivial from the definition of essential supremum but I'm unsure if this is what @stange meant.