Let be a function. First, we have the following trivial result:

Observation. If a non-negative funtion satisfies the following inequalityfor all , then we must have in .

The proof of the above observation depends on the non-negativity of . It is worth noting that we do not require the continuity of . Here in this post, we are interested in the following

Main result. If the non-negative, continuous function satisfies near zero andfor all , then there exists some in such a way that in .

The above result, in a special setting, appears in a recent work of Hyder and Sire. (See this if you cannot access the content.) Now we discuss a proof of the above main result, original due to one of my young colleagues.

**PROOF OF THE MAIN RESULT**

Fix a number in such a way that . In there is no such an , then we must have ., hence there is nothing to prove. Otherwise, we recursively define a sequence as follows

for . It is worth noting that in the above definition, the set of possible is not empty because at least belongs to this set. Now by the definition, we easily see that the sequence is non-negative and non-increasing. Hence there exists

with . There are two possibilities:

**Case 1**. Suppose that . In this case, we first have and then we have

for all . Now taking the limit as we easily get a contradiction because as .

**Case** 2. Suppose that . In this senario, we can take . Indeed, we shall prove that for all . By the definition we must have for all which implies

for all . Taking the limit as we obtain

for all $0 \leq x \leq x_\infty$. Here we use the continuity of . This and the hypothesis imply that

which immedialtely yields

Hence we also have for all . This shows that we can choose . The proof is complete.

**A POSSIBLE GENERALIZATION**

It appears that the hypothesis near is not mandatory. The only place that we need to make use of this is to rule out the possibility of . This is done because in a small neighborhood of zero the function is smaller than . Consequently, the above argument still works if we just assume that is not a global maximum point of on . Because of this, there would exist some point such that . Hence the argument works well.

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