Consider a continuous function $F:\Bbb R_{\ge0}\times\Bbb R_{\ge0}\setminus\Delta_{\Bbb R_{\ge0}^2}\to\Bbb R_{\ge0}$ where $\Delta_{\Bbb R_{\ge0}^2}:=\{(x,x):x\ge0\}$ is the diagonal of $\Bbb R_{\ge0}^2$.

Suppose moreover $F$ symmetric, that is $F(t,s)=F(s,t)\;\;\forall (s,t)\in\Bbb R_{\ge0}\times\Bbb R_{\ge0}\setminus\Delta_{\Bbb R_{\ge0}^2}$ and $F(0,t)>0$ for all $t>0$.

Is it true that $$ \lim_{T\to0+}\left[\sup_{0<t\le T}F(0,t)\right] =\lim_{T\to0+}\left[\sup_{0\le s<t\le T}F(s,t)\right]\;\;? $$ It seems clear that RHS is $\ge$ than LHS; is the converse true?

I have a proof, but in the comments there is a counterexample: where is the problem?

PROOF: We distinguish two cases for the value of RHS.

Let us suppose first $\lim_{T\to0+}\left[\sup_{0\le s<t\le T}F(s,t)\right]=c\in\Bbb R_{\ge0}$.

If $c=0$ then the conclusion is trivial. Suppose thus $c>0$.

Now for every $n\in\Bbb N$ there exists $\delta_n>0$ such that $$ \left|\sup_{0\le s<t\le\delta_n}F(s,t)-c\right|<\frac1n $$ and for every $n\in\Bbb N$, fixed $\delta_n$, there exists $\{(s_m^{(\delta_n)},t_m^{(\delta_n)})\}_{m\ge1}\subset\Bbb R_{\ge0}^2\setminus\Delta_{\Bbb R_{\ge0}^2}$ such that $$ F(s_m^{(\delta_n)},t_m^{(\delta_n)})\stackrel{m\to+\infty}{\longrightarrow} \sup_{0\le s<t\le\delta_n}F(s,t). $$ from which we get $$ \lim_{n\to+\infty}F(s_n^{(\delta_n)},t_n^{(\delta_n)}) =\lim_{T\to0+}\left[\sup_{0\le s<t\le T}F(s,t)\right]=c\; $$ (moreover it is clear that $s_n^{(\delta_n)},t_n^{(\delta_n)},\delta_n\to0$). Then, by continuity, $\forall\varepsilon>0\;\;\exists N_{\varepsilon}\ge1$ such that $$ \left|F(0,t_n^{(\delta_n)})-F(s_n^{(\delta_n)},t_n^{(\delta_n)})\right|<\frac{\varepsilon}2\;\;\;\;\;\forall n\ge N_{\varepsilon} $$ and $$ \left|F(s_n^{(\delta_n)},t_n^{(\delta_n)})-c\right|<\frac{\varepsilon}2\;\;\;\;\;\forall n\ge N_{\varepsilon}, $$ from which we get $$ \left|F(0,t_n^{(\delta_n)})-c\right|<\varepsilon\;\;\;\;\;\forall n\ge N_{\varepsilon}, $$ and thus $$ \lim_{T\to0+}\left[\sup_{0<t\le T}F(0,t)\right] =\lim_{n\to+\infty}\left[\sup_{0<t\le t_n^{(\delta_n)}}F(0,t)\right] \ge \lim_{n\to+\infty}F(0,t_n^{(\delta_n)})=c $$ which allows to conclude, since clearly $\lim_{T\to0+}\left[\sup_{0<t\le T}F(0,t)\right]\le c$.

The case $\lim_{T\to0+}\left[\sup_{0\le s<t\le T}F(s,t)\right]=+\infty$ is only slightly different: since $$ T\mapsto\sup_{0\le s<t\le T}F(s,t) $$ is not decresing, it is clear that $$ \sup_{0\le s<t\le T}F(s,t)=+\infty\;\;\;\forall T>0. $$ In particular fixing $n\ge1$ we have that $\sup_{0\le s<t\le \frac1n}F(s,t)=+\infty$, thus there exists $\{(s_m^{(n)},t_m^{(n)})\}_{m\ge1}\subset\Bbb R_{\ge0}^2\setminus\Delta_{\Bbb R_{\ge0}^2}$ such that $$ \lim_{m\to+\infty}F (s_m^{(n)},t_m^{(n)})=+\infty, $$ and since this holds true for every $n\ge1$ we can write \begin{equation} \lim_{n\to+\infty}F (s_n^{(n)},t_n^{(n)})=+\infty. \end{equation} Now, being clearly $s_n^{(n)},t_n^{(n)}\to0$ as $n\to+\infty$ and since $F$ is continuous, $\forall\varepsilon>0$ $\exists N_{\varepsilon}\ge1$ such that $$ \left|F(0,t_n^{(n)})-F(s_n^{(n)},t_n^{(n)})\right|<\varepsilon\;\;\;\;\;\forall n\ge N_{\varepsilon}; $$ in particular we get $$ F(0,t_n^{(n)})>F(s_n^{(n)},t_n^{(n)})-\varepsilon\;\;\;\;\;\forall n\ge N_{\varepsilon}, $$ from which $$ \lim_{n\to+\infty}F (0,t_n^{(n)})=+\infty. $$ Then we can conclude as in the previous case: $$ \lim_{T\to0+}\left[\sup_{0<t\le T}F(0,t)\right] =\lim_{n\to+\infty}\left[\sup_{0<t\le t_n^{(n)}}F(0,t)\right] \ge \lim_{n\to+\infty}F(0,t_n^{(n)})=+\infty. $$

FINALLY: I wanted to treat my problem in general, but originally it was $$ F(s,t)=\frac{|f(t)-f(s)|}{|t-s|^{\lambda}} $$ for some $0<\lambda\le1$ and for a function $f$ which is $\alpha$-Holder continuous for some $\alpha$.

THE MOST IMPORTANT THING: If my proof is wrong, what kind of hypothesis we can put in order to obtain the result?

THE GENESIS OF THIS POST: $f$ originally was the difference of two solutions of the following scalar SDE: $$ x_t=\underbrace{\xi_0+\int_0^tb(s,x_s)\,ds+\int_0^t\sigma(s,x_s)\,dW_s^H}_{=:z_t}+y_t\; $$ where $$ y_t=\sup_{s\in[0,t]}\left(z_s\right)^{-}. $$ and $$ b:\Bbb R_{\ge0}\times\Bbb R\to\Bbb R $$ and $$ \sigma:\Bbb R_{\ge0}\times\Bbb R\to\Bbb R $$ are bounded measurable functions, called drift and diffusion coefficient respectively. We will assume the following hypotesis on them:

$$ |b(t,x)-b(t,y)|\le K_0|x-y|\;\;\forall x,y\in\Bbb R,\;\forall t\in[0,T] $$

$$ |\sigma(t,x)-\sigma(t,y)|\le K_0|x-y|\;\;\forall x,y\in\Bbb R,\;\forall t\in[0,T] $$

$$ |\sigma(t,x)-\sigma(s,x)|\le K_0|t-s|^{\nu}\;\;\forall x\in\Bbb R,\;\forall s,t\in[0,T] $$ and $W^H$ is the fractional Brownian motion of Hurst index $1/2<H<1$; moreover the integral is the YOUNG INTEGRAL, which is defined as the limit of the usual Riemann Sums, or it can be equivalently viewed as follows (here we deal with deterministic case: we think to $W^H$ as a fixed path of the fBM): if $h$ is a $\lambda$-Holder real function, for $\lambda>1-H$, then it can be proved that $$ \int_s^th_u\,dW_u^H=h_s(W_t^H-W_s^H)+\Lambda_{st}((h_c-h_a)(W_b^H-W_c^H)) $$ where $a,c,b$ are mute variables and $\Lambda:\mathcal{ZC}_3^{\mu}\to\mathcal{C}_2^{\mu}$ is the inverse of $\delta:\mathcal{C}_2^{\mu}\to\mathcal{ZC}_3^{\mu}$ which is defined as $$ (\delta h)_{sut}:=-h_{ut}+h_{st}-h_{su} $$ and $$ \mathscr C_2^{\mu}:=\{h\in\mathscr{C}_{2}\;:\;\|h\|_{\mu}<+\infty\} $$ where $\mathscr{C}_{2}$ is the $\Bbb R$-vector space of all functions $h:[0,T]^2\to\Bbb R$ continuous and such that they vanish on diagonal. Moreover $$ \|h\|_{\mu}:=\sup_{\substack{s,t\in[0,T]\\s\neq t}}\frac{|h_{ts}|}{|t-s|^{\mu}} $$ and $$ \mathscr C_3^{\mu}:=\{h\in\mathscr{C}_{3}\;:\;\|h\|_{\mu}<+\infty\} $$ where $\mathscr{C}_{3}$ is the $\Bbb R$-vector space of all functions $h:[0,T]^3\to\Bbb R$ continuous such that $h_{sut}=0$ whenever $s=u$ or $u=t$ and $\|\cdot\|_{\mu}$ is a suitable norm.

Thus \begin{align*} f_t &=x_t^{(1)}-x_t^{(2)}\\ &=\int_0^t(b(u,x_u^{(1)})-b(u,x_u^{(2)}))\,du+ \int_0^t(\sigma(u,x_u^{(1)})-\sigma(u,x_u^{(2)}))\,dW_u^H +y_t^{(1)}-y_t^{(2)} \end{align*}

and $$ F(0,t)=\frac{|f_t|}{t^H} $$ IMPORTANT since the original exponent is $\rho>1$, then it would be enough to prove that this last limit is nonzero, in fact $$ \frac{|f_t|}{t^{\rho}}=\frac{|f_t|}{t^{H}}\frac1{t^{\rho-H}} $$

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    $\begingroup$ Is the point $(0,0)$ excluded from the domain? If, yes I would try the following F with $F(0,t) = 0 $ with $t \ge 0 $, $ F(s,t) = 1$ for $s=t$ and else is between $0$ and $1$ (and make F continuous). That should gurantee LHS $\neq$ RHS. Is my approach too naive? If $(0,0)$ is included, it looks like that due to a maximum has to be obtained for a continuous function on a compact set. $\endgroup$
    – Imago
    Commented Sep 10, 2017 at 19:19
  • $\begingroup$ Many thanks for the comment! Now $(0,0)$ is excluded, since it belongs to the diagonal of $\Bbb R_{\ge0}^2$. However since $F$ is NOT defined on the diagonal, defining $F(s,t)=1$ for $s=t$ sounds senseless. Do you agree? $\endgroup$
    – Joe
    Commented Sep 10, 2017 at 19:25
  • $\begingroup$ Either way, aren't you just looking at its supremum? I just wanted to make sure that $F(s,t) \le 1$ for $0 \le s \lt t $ $\endgroup$
    – Imago
    Commented Sep 10, 2017 at 19:28
  • $\begingroup$ @Imago: I missed the fact $F$ has to be symmetric. $\endgroup$
    – Joe
    Commented Sep 10, 2017 at 20:21
  • $\begingroup$ I don't see how this changes something, if you keep the LHS constant $0$ and the RHS constant $\gt 0$ - I imagine the sup to be constantly achieved somewhere for $0 \le s \lt t $ and else fitting in the points to preserve F's continuity. Imo, the symmetry of F is rather unimportant, not too relevant for this exercise. $\endgroup$
    – Imago
    Commented Sep 10, 2017 at 20:40

2 Answers 2


The statement "Then by continuity, $\forall \epsilon>0$, $\exists N_\epsilon$..." is wrong. If a function $F$ is continuous on an open set and $\xi_n$ is a sequence converging to a boundary point then $F(\xi_n)$ need not be Cauchy (which in fact is the claim you make). For the argument to hold you need uniform continuity, but then $F$ extends continuously to $(0,0)$ which is not what you are interested in. A concrete example is: $ F(s,t) = \frac{s}{t}$ for $0\leq s<t$ and $F(s,t) = \frac{t}{s}$ for $0\leq t < s$. You have $\limsup_T F(0,T)=0$ and $\limsup_{s,t} F(s,t)=1$.

The application you have in mind is, however, somewhat different. So let $$ F(s,t) = \frac{f(s)-f(t)}{|s-t|^\lambda}$$ with $f$ an $\alpha$-Hölder continuous function.

If $ 0 < \lambda < \alpha \leq 1 $ then $F$ is in fact uniformly continuous, since you have: $$ |F(s,t) | \leq \frac{ C|t-s|^\alpha}{|t-s|^\lambda} = C |t-s|^{\alpha-\lambda}.$$ so there is no problem.

In the case $0<\lambda=\alpha\leq 1$ the conclusion is, however, wrong. To constuct a counter-example, fix $k>0$ and define for $A>0$ the spike of amplitude $A$: $$ \phi_A(x) = \max\{0, A - k |x|^\alpha\}$$ This non-negative function is bounded by $A$, has support in $|x|\leq (A/k)^{1/\lambda}$ and has $\alpha$-Hölder constant $k$ (independent of $A$): $$ \sup_{x\neq y} \frac{|\phi_A(x)-\phi_A(y)|}{|x-y|^\alpha} = k .$$

Now, define $$ f(t) = \sum_{n\geq 1} \phi_{\frac{1}{n^{2\lambda}}} (t-\frac{1}{n}) .$$ The spikes have disjoint support for $k$ large enough and $f(1/n)=1/n^{2\lambda}$ so the function verifies: $f(t)\leq t^{2\lambda}$, $t\geq 0$. In particular, $$ \limsup_{T\rightarrow 0^+} F(0,T) = \limsup_T f(T)/T^\lambda \leq C \limsup_T T^{2\lambda-\lambda} = 0. $$ On the other hand we have when $\alpha=\lambda$: $$ \limsup_{T\rightarrow 0^+} \sup_{0<t<s\leq T} F(t,s)= \limsup_T \sup_{0<t<s\leq T} \frac{|f(s)-f(t))|}{|s-t|^\lambda} = k $$ since arbitrarily close to the origin we have a spike with Hölder constant $k$. By letting $k$ increase with $n$ you may also construct an example where the first limsup is zero and the last infinity. In other words in order to obtain the result you need better control on the function $f$ when approaching the origin.

According to one of your comments elsewhere on this page it looks as if you may be interested in the case $\alpha=\lambda=1/2$ and $f(t)$ being e.g. a standard Brownian motion. In that case I believe that a.s. both limits are infinity (so in a sense you are then right). It is well-know that the $\limsup_{T}\sup_{s,t}$ is infinity. This follows for example from Lévy's modulus of continuity. see e.g. Peres and Morters, Thm 1.14. I believe that is also true a.s. for the point wise limit $\limsup_T F(0,T)$ but couldn't find a direct proof. There exists points where the limit is finite (sometimes called $\alpha$ slow times) but it seems that they have zero Lebesgue measure, although possibly Hausdorff dim 1 (?). But I am far from sure about this last point.

  • $\begingroup$ First of all many thanks for your answer. Then, I am dealing with fractional Brownian motion of index $0<H<1$, but things don't change. All started because I wanted to prove that $\limsup_{T\to0^+}\frac{|f_T-f_0|}{|T-0|^H}=+\infty$,and I thought it was easier to pass thru the equality which is the object of this post, but it isn't necessary:the most important thing is to prove this last limit is indeed $=+\infty$.Moreover the last remark is that $f$ is not the fBM, but a map which involves the fBM.And this is not easy, however, I am gonna think to extend the post explaining the whole problem $\endgroup$
    – Joe
    Commented Sep 24, 2017 at 12:28
  • $\begingroup$ If you are solving a stochastic ode satisfying Lipschitz conditions with fBM driving term, I think it is just the behavior of the fBM that will be reflected in the ode solution (as far as I recall that is the case with BM). As I mentioned the pointwise limsup=infinity is more delicate than the two variables limsup. That is probably also the case with fBM. You could perhaps consider posting a more detailed and precise question on MO rather than SE (going directly to the relevant thing you want to prove)? Anyway, the question is interesting. $\endgroup$
    – H. H. Rugh
    Commented Sep 24, 2017 at 12:40
  • $\begingroup$ "I think the behavior of the fBM is reflected in the solution": This is EXACTLY what I've thought, and that is the reason I have begun to work in this direction: if it was true, than $\limsup_{T\to0+}\frac{|f_T|}{T^H}=+\infty$ and I would have reached the conclusion I hoped. However I have expanded the post in order to make understand what did lead me here. Moreover: thanks again $\endgroup$
    – Joe
    Commented Sep 24, 2017 at 13:44
  • $\begingroup$ Glad that my response might help. You might however review the way you use the bounty system. You have posed a question to which I seem to have given a full response which seemingly have taken you off a wrong track for you phd thesis. Glad to hear this. But you have now added to your original question an extended part which seems to reflect a central part of your thesis. This is an important point to you and I understand your motive but I don't think it reflects the policy of the use of bounty's on SE. It would be more courteous to pose a question,get an answer, close it and pose another. $\endgroup$
    – H. H. Rugh
    Commented Sep 25, 2017 at 19:07
  • $\begingroup$ This being said, I see on relevant posts on MO that LIL argument pretends that the limit of $f(t)/t^H$ is $+\infty$. As I mentioned I think it is not obvious that this is right a.e. I don't understand your exponent $\rho$? If $\rho>1$ then I think it is quite automatic that $f(t)/t^\rho$ goes to infinity a.e.? $\endgroup$
    – H. H. Rugh
    Commented Sep 25, 2017 at 20:16

I have a proof, but in the comments there is a counterexample: where is the problem?

The problem is here

we get $$ \lim_{n\to+\infty}F(s_n^{(\delta_n)},t_n^{(\delta_n)}) =\lim_{T\to0+}\left[\sup_{0\le s<t\le T}F(s,t)\right]=c\; $$

The convergence for each fixed $n$ of a sequence $a_{m,n}$ to a limit $c_n$ when $m$ tends to infinity does not necessarily implies that a diagonal sequence $a_{n,n}$ converges to a $\lim c_n$ when $n$ tends to infinity. For instance, we can put $a_{m,n}=0$ when $m=n$ and $a_{m,n}=1$, otherwise.

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    $\begingroup$ Thanks for the answer. Do you know some suitable condition under which, the claim is true? $\endgroup$
    – Joe
    Commented Sep 23, 2017 at 17:21
  • $\begingroup$ @Joe Unfortunately, I don’t know such non-trivial conditions. Are the counterexamples exist also for your original problem about $\alpha$-Hölder continuous functions? $\endgroup$ Commented Sep 23, 2017 at 18:20
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    $\begingroup$ my problem arises from a paper about a stochastic differential equation, in which existence was proved, but not uniqueness. Then my master degree thesis is trying to prove uniqueness. So I am working on this, and I found a way, but it passes thru the property I stated in this post. Hence it is a "new way" and this a "new land" for me, thus I don't know so much about counterexamples. We are dealing with Brownian motion, so the paths are "constantly irregular", but stating the whole problem would be too too too long $\endgroup$
    – Joe
    Commented Sep 23, 2017 at 19:30
  • $\begingroup$ @Joe OK, I'll think about your original problem. I don't need for that that differential equation staff. $\endgroup$ Commented Sep 23, 2017 at 19:57
  • $\begingroup$ many thanks! I have begun to work on this problem before the beginning of the summer and now I'm really tired! $\endgroup$
    – Joe
    Commented Sep 23, 2017 at 19:59

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