Ngô Quốc Anh

July 24, 2021

A contraction type argument

Filed under: Uncategorized — Ngô Quốc Anh @ 10:47

Let f : [0,1] \to [0, +\infty) be a function. First, we have the following trivial result:

Observation. If a non-negative funtion f satisfies the following inequality

\displaystyle f(x) \leq \frac 12 f(x)

for all 0 \leq x \leq 1, then we must have \displaystyle f \equiv 0 in [0,1].

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

Main result. If the non-negative, continuous function f satisfies f=o(1) near zero and

\displaystyle f(x) \leq \frac 12 \sup_{0 \leq y \leq x} f(y)

for all 0 \leq x \leq 1, then there exists some \delta >0 in such a way that \displaystyle f \equiv 0 in [0,\delta].

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.

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