Let $C_p\equiv C_p(\mathbb R_+,\mathbb R_+)$ be the space of left-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 $n+1$.

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 $n+1$.

Let $C_p\equiv C_p(\mathbb R_+,\mathbb R_+)$ be the space of left-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 $1\wedge \tau:=\min(1,\tau)$.

Let $C_p\equiv C_p(\mathbb R_+,\mathbb R_+)$ be the space of left-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 $n+1$.