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Let $C_p\equiv C_p(\mathbb R_+,\mathbb R_+)$ be the space of right-continuous piecewise constant functions $f: \mathbb R_+\to \mathbb R_+$, i.e. $f\in C_p$ iff

$$f(t)=\sum_{k=1}^n {\mathbf 1}_{[t_{k-1},t_k)}(t)x_k,$$

where $0=t_0<t_1<\cdots<t_{n-1} <t_n=\infty$, $x_1,\ldots, x_n\in (0,\infty)$ and $n\ge 1$.

For any $f\in C_p$ and any $t>0$, define respectively

$$\Phi_f(s):=2\int_0^s f(u)^2du; \quad \tau_f(t):=\inf\big\{s\ge 0: \Phi_f(s)\ge t\big\}.$$

Can we find some $c>0$ (e.g. I believe $c=1/e$) such that for any $T>0$ and any $f\in C_p$ the following inequality holds :

$$\int_0^{1\wedge \tau} \big(1+\log(f(s))\big)ds \le \int_0^{1\wedge \tilde\tau} \big(1+\log(\tilde f(s))\big)ds,$$

where $\tilde f(s):=\max(f(s),c)$, $\tau:=\tau_f(T)$, $\tilde\tau:=\tau_{\tilde f}(T)$ and $1\wedge \tau:=\min(1,\tau)$.

PS : By definition $\tilde \tau\le \tau$. A natural attempt is to argue by induction. The case $n=1$ is trivial. For $n=2$, with a straightforward but heavy verification by distinguishing different cases $\tilde \tau< \tau\le 1$ and $\tilde \tau\le 1<\tau$, the desired inequality holds. But I don't know how to proceed from general $n$ to $n+1$.

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It takes some time to understand what is asked, but after that it becomes easy.

Suppose that $\newcommand{\wt}{\widetilde}$ $+\infty>\tau\ge \wt\tau\ge 0$ and $\int_0^\tau F\le\int_0^{\wt\tau}\wt F$ where $F=\max(F,e^{-2})$ (so my $F$ is your $f^2$). Write $F=e^{-2}+G$ and decompose $G=G_+-G_-$ where $G_\pm=\max(\pm G,0)$.

What we are given is $$ \int_0^\tau (e^{-2}+G_+-G_-)\le\int_0^{\wt\tau}(e^{-2}+G_+)\,, $$ i.e., $$ e^{-2}(\tau-\wt\tau)+\int_{\wt\tau}^\tau G_+-\int_0^\tau G_-\le 0. $$

What we want is $$ \int_0^\tau [1+\tfrac 12\log(e^{-2}+G_+-G_-)]\le\int_0^{\wt\tau}[1+\tfrac12\log(e^{-2}+G_+]\,, $$ Since $G_+$ and $G_-$ have disjoint supports and $1+\frac 12\log(e^{-2})=0$, we have $$ 1+\tfrac 12\log(e^{-2}+G_+-G_-)= [1+\tfrac 12\log(e^{-2}+G_+)]+[1+\tfrac 12\log(e^{-2}-G_-)] $$ so we can also rewrite it as $$ \int_{\wt\tau}^\tau[1+\tfrac 12\log(e^{-2}+G_+)] +\int_0^\tau[1+\tfrac 12\log(e^{-2}-G_-)\le 0\,. $$ However $x\mapsto 1+\frac 12\log(e^{-2}+x)$ is concave, vanishes at $0$, and its derivative $e^2$ at $0$ is positive, so linearizing at $0$, we see that it suffices to prove that $$ \int_{\wt\tau}^\tau G_+ -\int_0^\tau G_-\le 0\,, $$ which is weaker than what is given.

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  • $\begingroup$ Thank you very much fedja for your answer. I believe that I need to read very carefully your reasoning, and please allow me to come back towards you if it's not clear to me $\endgroup$
    – Fawen90
    Commented Apr 3, 2023 at 14:01
  • $\begingroup$ @Fawen90 Yeah, ask as many questions as you want :-) $\endgroup$
    – fedja
    Commented Apr 3, 2023 at 16:14
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    $\begingroup$ Fantastic reasoning! Indeed you have proved a stronger result for all measurable and bounded functions $f:\mathbb R_+\to\mathbb R_+$. This is exactly what I need. Just one typo, $\widetilde F=\max(F,e^{-2})$ instead of $F=\max(F,e^{-2})$. Also, I think we should have equality $\int_0^\tau F=\int_0^{\widetilde\tau}\widetilde F$. Do you think so? $\endgroup$
    – Fawen90
    Commented Apr 3, 2023 at 20:29
  • $\begingroup$ Thanks again for your help. I do appreciate your trick $\endgroup$
    – Fawen90
    Commented Apr 3, 2023 at 20:30
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    $\begingroup$ @Fawen90 1) We'll have equality if $\tau\le 1$. If $1$ is in between, then $\tau\wedge 1$ becomes the $\tau$ I'll use in my argument and then we only have an inequality. The case $\widetilde\tau>1$ is trivial. 2) Yep, it should be $\widetilde F$. $\endgroup$
    – fedja
    Commented Apr 3, 2023 at 20:43

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