# 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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