Understanding Continuous Functions and Their Properties: A Deep Dive into the Problem of ( f(x) f(2x) )

Continuous functions play a crucial role in mathematical analysis, and the problem of determining when ( f(x) f(2x) ) holds true for some ( x ) within a specified interval is one such intriguing case. The article delves into a series of simplified versions of this problem, which can help in understanding and dissecting the underlying mathematical principles.

Baby Version 1

Consider a continuous function ( f: [frac{1}{2}a, a] to mathbb{R} ) with the conditions that ( f(1) f(a) ) and ( f(x) eq f(1) ) for all ( 1 le x

Proof

To solve this, we define a new function ( h(x) f(2x) - f(x) ). Notice that the domain of ( h(x) ) is the interval ( frac{1}{2}a le x le a ).

First, evaluate ( h(x) ) at the endpoints: ( h(frac{1}{2}a) f(a) - f(frac{1}{2}a) ) ( h(a) f(2a) - f(a) )

Since ( f(1) f(a) ) and ( f(x) eq f(1) ) for all ( 1

Baby Version 2

In this version, the function ( f: [frac{1}{4}, 1] to mathbb{R} ) satisfies ( f(frac{1}{4}) f(1) ). We need to prove that there is a point ( x ) in the interval ( frac{1}{4} le x le 1 ) such that ( f(x) f(2x) ).

Proof

Define ( h(x) f(2x) - f(x) ). At the endpoints, we have:

( h(frac{1}{4}) f(frac{1}{2}) - f(frac{1}{4}) ) ( h(1) f(2) - f(1) )

Given ( f(frac{1}{4}) f(1) ), if ( f(frac{1}{2}) eq f(frac{1}{4}) ) and ( f(2) eq f(1) ), the values of ( h(frac{1}{4}) ) and ( h(1) ) will be such that they will have opposite signs, ensuring the existence of a point ( x ) where ( h(x) 0 ).

Baby Version 3

For the interval ( f: [frac{1}{2^n}, 2^{n-1}] to mathbb{R} ) with ( f(frac{1}{2^n}) f(2^{n-1}) ), we aim to show that there must be a point ( x ) in ( frac{1}{2^n} le x le 2^{n-1} ) such that ( f(x) f(2x) ).

Proof

We use the function ( h(x) f(2x) - f(x) ). At specific points, the product of differences is zero:

( h(frac{1}{2^n}) cdot h(frac{1}{2^{n-1}}) cdot h(frac{1}{2^{n-2}}) cdots h(2^{n-1}) 0 ) Since the function ( h(x) ) is a continuous function on the closed interval, if the product of these values is zero, there must be a root in the interval.

Hence, there exists a point ( x ) in the specified interval where ( h(x) 0 ), that is, ( f(x) f(2x) ).

Better Version

The question at hand is to find all real numbers ( a ge 1 ) for which any continuous function ( f: [1, 2a] to mathbb{R} ) satisfying ( f(1) f(2a) ) has a point ( x ) in the interval ( 1 le x le a ) such that ( f(x) f(2x) ).

Conclusion

Based on the analysis of the simplified versions, we can observe that the condition ( a 5 ) does not satisfy the problem’s requirements. The intermediate value theorem and the properties of continuous functions form the basis of the proof in all instances, thereby confirming that certain values of ( a ) indeed satisfy the condition.