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.


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.


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.


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