Let $C$ be the set of continuous functions on $[0,T]$ taking values in $[0,1]$. Denote $\|f-g\|_t:=\max_{0\le u\le t}|f(u)-g(u)|$ for $f,g \in C$ and $t\in [0, T]$. Let $\phi: C\times C\times \{(s,t): 0\le s<t\le T\}\to\mathbb R_+$ be a measurable function s.t.

$$\Phi(f,g,t):=\int_0^t\phi(f,g,s,t)ds \quad \le \quad \sqrt{t},\quad \forall f,g\in C \mbox{ and } t\in [0,T].$$

If $\phi(f,g,s,t)\le \|f-g\|_t\big/(t-s)$ for all $f,g\in C$ and $0\le s<t\le T$, can we prove (or find a counterexample), for some constant $L>0$,

$$\Phi(f,g,t)\quad\le\quad L\sqrt{t}\|f-g\|_t,\quad \forall f,g\in C \mbox{ and } t\in[0, T]?$$

PS : It might be useful to give an alternative formulation (according to my context). Indeed, $\phi$ appearing above satisfies

\begin{eqnarray} \phi(f,g,s,t) &\le& \frac{1}{\sqrt{t-s}},\quad \forall f,g\in C \mbox{ and } 0\le s<t\le T\\ \phi(f,g,s,t) &\le& \frac{\|f-g\|_t}{t-s},\quad \forall f,g\in C \mbox{ and } 0\le s<t\le T. \end{eqnarray}

Can we deduce, for some $L>0$ and $a \in [1/2,1)$,

$$\phi(f,g,s,t) \quad \le\quad \frac{L\|f-g\|_t}{(t-s)^a},\quad \forall f,g\in C \mbox{ and } 0\le s<t\le T?$$


1 Answer 1


This is not true. Take

$$\phi(f,g,s,t):=\min\left(\frac {\|f-g\|_t}{t-s},\frac{1}{\sqrt{t-s}}\right).$$

Then for any $f,g\in C$ with $\delta:=\|f-g\|_t \le \sqrt{t}$, it holds that

$$\Phi(f,g,t)=\int_0^{t-\delta^2}\frac {\|f-g\|_t}{t-s} +\int_{t-\delta^2}^t\frac{1}{\sqrt{t-s}}=\delta\left(\log\left(\frac{t}{\delta^2}\right)+2\right),$$

which violates your desired inequality.

  • $\begingroup$ Many thanks for the example $\endgroup$
    – GJC20
    Apr 23, 2022 at 18:39

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