# 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.

PROOF OF THE MAIN RESULT

Fix a number $x_0 \in (0,1]$ in such a way that $f(x_0) > 0$. In there is no such an $x_0$, then we must have $f \equiv 0$., hence there is nothing to prove. Otherwise, we recursively define a sequence $(x_k)_{k=0}^\infty$ as follows

$\displaystyle x_{k+1} = \inf \Big\{ y \in [0,x_k] : f(y) \geq f(x_k) \Big\}$

for $k\geq 0$. It is worth noting that in the above definition, the set of possible $y$ is not empty because at least $x_k$ belongs to this set. Now by the definition, we easily see that the sequence $(x_k)_k$ is non-negative and non-increasing. Hence there exists

$\displaystyle x_\infty = \lim_{k \nearrow \infty} x_k$

with $x_\infty \in [0, x_0]$. There are two possibilities:

Case 1. Suppose that $x_\infty = 0$. In this case, we first have $x_k \searrow 0$ and then we have

$f(x_{k+1}) \geq f(x_k) \geq \cdots \geq f(x_0) > 0$

for all $k$. Now taking the limit as $k \nearrow \infty$ we easily get a contradiction because $f(x_k) = o(1)$ as $k \nearrow \infty$.

Case 2. Suppose that $x_\infty \ne 0$. In this senario, we can take $\delta = x_\infty$. Indeed, we shall prove that $f(x) =0$ for all $x \in [0, \delta]$. By the definition we must have $x \leq x_k$ for all $k$ which implies

$f(x) \leq f(x_k)$

for all $k$. Taking the limit as $k \nearrow \infty$ we obtain

$f(x) \leq f(x_\infty)$

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

$\displaystyle f(x_\infty) \leq \frac 12 \sup_{0 \leq y \leq x_\infty} f(y) \leq \frac 12 f(x_\infty),$

which immedialtely yields

$f(x_\infty) = 0.$

Hence we also have $f(x) = 0$ for all $0 \leq x \leq x_\infty$. This shows that we can choose $\delta =x_\infty$. The proof is complete.

A POSSIBLE GENERALIZATION

It appears that the hypothesis $f=o(1)$ near $0$ is not mandatory. The only place that we need to make use of this is to rule out the possibility of $x_\infty = 0$. This is done because in a small neighborhood of zero the function $f$ is smaller than $f(x_0)$. Consequently, the above argument still works if we just assume that $0$ is not a global maximum point of $f$ on $[0,1]$. Because of this, there would exist some point $x_0 \in (0,1]$ such that $f(x_0) > f(0)$. Hence the argument works well.