I am reading the paper "The long-time behaviour of a stochastic SIR epidemic model with distributed delay and multidimensional Levy jumps" by Driss Kiouach and Yassine Sabbar.

I have two questions detailed below regarding the proof of Theorem 3.7 in the paper (so reading the entire paper is not required):

  1. On page 8, equation (10) is stated as: $$ \int_\mathcal{U}\Bigg[\ln\Bigg(1+\frac{\omega_1 \lambda_2(u)I(t)+\omega_2 \lambda_4(u)D(t)}{\omega_1 I(t)+\omega_2 D(t)} \Bigg)-\frac{\omega_1 \lambda_2(u)I(t)+\omega_2 \lambda_4(u)D(t)}{\omega_1 I(t)+\omega_2 D(t)} \Bigg) \Bigg]\nu(du)\leq \Pi, $$ where $\Pi$ is defined at the bottom of page 7. How did the authors invoke this inequality?

  2. Hölder's inequality. If we go to page 9 in the pdf, the authors make use of Hölder's inequality: $$ t^{-1}\int_0^t \Bigg|\psi(s)-\frac{A}{\mu_1}\Bigg|ds \leq t^{-\frac{1}{2}}\Bigg(\int_0^t \Big(\psi(s)-\frac{A}{\mu_1}\Big)^2\Bigg)^{1/2}= \cdots $$ The application of Hölder's inequality inside the integral part is clear. However I'm not sure on how it was applied to $t^{-1}$ to obtain $t^{-\frac{1}{2}}$: it seems trivial but I just cant see it, any ideas?

  • 2
    $\begingroup$ I don't understand the downvote, I believe my question is clear. $\endgroup$
    – Math
    Oct 4 at 15:06
  • 4
    $\begingroup$ Probably because this question does not fit here well. It would be better suited for Math.SE $\endgroup$
    – Max Horn
    Oct 4 at 21:33

1 Answer 1


Some general principles with these sorts of "how did this step get justified" questions:

  1. Try to identify the key mathematical objects (variables, expressions, functions, etc.) involved and give them temporary names (if they don't already have one). Moderately complicated objects that appear two or more times in the statement to be justified are particularly prime candidates for being assigned a temporary name.
  2. Abstract away any information that you suspect to be irrelevant (see item 10 of this blog post of mine for further elaboration of this point). It's OK if your initial abstraction turns out to fail - the counterexample often gives very instructive clues as to how to utilize the information that you previously believed to be irrelevant.
  3. Make a guess as to how the justification might proceed, and in particular identify some simpler subclaim that could imply the original claim. Again, it's OK if your initial guess fails - the failure again will likely provide instructive clues as to what the alternative should be.
  4. Above all, try something. Staring at a problem and declaring that you have no idea how to proceed is what Gowers in this blog post refers to as having a "fake difficulty". As stated in Principles 2 and 3, at this stage of the problem solving process one should not be afraid to make guesses that fail; this is far more productive than not guessing at all. (Nobody is going to administer an electric shock to you if you guess incorrectly.)

Now let's return to Question 1 and apply some of the above principles. The expression $\frac{\omega_1 \lambda_2(u) I(t) + \omega_2 \lambda_4(u) D(t)}{\omega_1 I(t) + \omega_2 D(t)}$ shows up twice here, so, following Principle 1, let's give it a name, say $X(u)$: $$ X(u) := \frac{\omega_1 \lambda_2(u) I(t) + \omega_2 \lambda_4(u) D(t)}{\omega_1 I(t) + \omega_2 D(t)}.$$ (I include a dependence on $u$ here because we will be integrating in $u$, but since we aren't doing anything interesting in the $t$ variable, I am simplifying the notation by dropping the $t$ dependence.) Also, if one looks at both the inequality to be justified, as well as the definition of $\Pi$, then the function $x \mapsto \log(1+x) - x$ shows up repeatedly, so let's give that a name too: $$ f(x) := \log(1+x)-x.$$ Our task is now to show that $$ \int_{\mathcal U} f(X(u))\ \nu(du) \leq \Pi \tag{1}$$ where from the article, $\Pi$ is defined (after concatenating some definitions and using our new function $f$) as $$ \int_{\mathcal U} f(\lambda_2(u) \wedge \lambda_4(u)) 1_{\lambda_2(u) \wedge \lambda_4(u) > 0} + f(\lambda_2(u) \vee \lambda_4(u)) 1_{\lambda_2(u) \vee \lambda_4(u) \leq 0}\ \nu(du).$$

Comparing both sides of (1) we see that both sides are integrals over the same domain ${\mathcal U}$ using the same measure $\nu(du)$. So, as per Principle 3, we can make a guess that we are supposed to establish a pointwise inequality $$ f(X(u)) \leq f(\lambda_2(u) \wedge \lambda_4(u)) 1_{\lambda_2(u) \wedge \lambda_4(u) > 0} + f(\lambda_2(u) \vee \lambda_4(u)) 1_{\lambda_2(u) \vee \lambda_4(u) \leq 0}$$ for each $u$ separately (as opposed to, say, using some integral identity such as the integration by parts formula). Let us now make this guess, and simplify the notation further by omitting the $u$ dependence, so we now want to show $$ f(X) \leq f(\lambda_2 \wedge \lambda_4) 1_{\lambda_2 \wedge \lambda_4 > 0} + f(\lambda_2 \vee \lambda_4) 1_{\lambda_2 \vee \lambda_4 \leq 0}. \tag{2}$$ We again make a guess (Principle 3): this should be justified by combining some property (such as positivity, monotonicity, or convexity) of the function $f$ with some relationship between $X$, $\lambda_2$, and $\lambda_4$ (as opposed to exploiting some relationship between $f$ and the arguments $X$, $\lambda_2 \wedge \lambda_4$, $\lambda_2 \vee \lambda_4$). Let's set $f$ aside for now and take a closer look at $X$. Invoking Principle 1 again, we temporarily introduce some more notation $$ a := \omega_1 I(t); \quad b := \omega_2 D(t)$$ and now $X$ can be written as $$ X = \frac{a \lambda_2 + b \lambda_4}{a+b}.$$ Invoking Principle 2, we suspect that the only properties of $a$ and $b$ that are likely to be relevant here are that they are positive real numbers. At this point it should be clear the relationship between $X$, $\lambda_2$, and $\lambda_4$: the former quantity is a convex combination of the latter two! So we are reduced to the task of establishing the inequality (2) whenever $X$ is a convex combination of $\lambda_2$ and $\lambda_4$.

At this point it is all up to the function $f$. I recommend that you plot this function (or get some computer algebra package to plot it for you), and try to prove this inequality manually. Given the presence of the cutoffs $1_{\lambda_2 \wedge \lambda_4 > 0}$ and $1_{\lambda_2 \vee \lambda_4 \leq 0}$, it will make sense to split into cases depending on the sign of $\lambda_2$ and $\lambda_4$, and maybe also to split into cases depending on which of $\lambda_2$ and $\lambda_4$ is larger. (One may end up initially splitting up into more cases than is absolutely necessary, but that is OK - the initial justification is often a bit messier than needed, and can always be tidied up later. At this stage of the problem solving process, it is more important to find an argument that works than to find an argument that is elegant.)

I'll let you work out your second question on your own using these principles. Some hints to get you started: as per Principle 1, one can introduce the function $$ g(s) := \psi(s) - \frac{A}{\mu_1}.$$ One can guess that the $t^{-1}$ factor is going to be cancelled out, so that one is actually trying to prove $$ \int_0^t |g(s)|\ ds \leq t^{1/2} (\int_0^t g(s)^2\ ds)^{1/2}.$$ Holder's inequality involves integrals, so the factor $t$ here is presumably going to come from an integral identity such as $t = \int_0^t 1\ ds$. Now you should be able to do the rest on your own.

  • 34
    $\begingroup$ I wish there were a heart sign instead of just upvoting. Here, Terry has taken the time and effort to give instruction on how to proceed, in the face of a possibly overcomplicated question with loads of symbols and a mismatched parenthesis, that most of us might not even get to the end of reading. $\endgroup$ Oct 4 at 19:20
  • $\begingroup$ What did the authors mean by " to find an optimal and good majorization"? $\endgroup$
    – Leo
    Oct 9 at 11:22

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