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 $n+1$.



An integral on the interval depending on the integrand
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.


