Question

Let f : [1,[infinity][→ Rd be a continuous function such that f(2x) = f(x) for all x ∈ [1,[infinity][.

1. Give an example of a nonconstant function verifying this property (you can
take d = 1).
2. Show that the function f is fully determinate by its restriction to [1, 2]. That is, if g is another function such that g(2x) = g(x) and g(x) = f(x) for all
x ∈ [1, 2] then g(x) = f(x) for all x ∈ [1,[infinity][.
3. Show that any function f v'erifying these properties is uniformly continuous.

attached the exercise in french

Answer 1

1. An example of a nonconstant function that satisfies the given property is f(x) = sin(2πlog_2(x)). 2. The function f is fully determinate by its restriction to [1, 2]. 3. Any function satisfying these properties is uniformly continuous.

Step-by-step explanation:

1. An example of a nonconstant function that satisfies the given property is f(x) = sin(2πlog_2(x)). This function is continuous and satisfies f(2x) = f(x), since sin(2πlog_2(2x)) = sin(2πlog_2(x) + 1) = sin(2πlog_2(x)).

2. To prove that the function f is fully determinate by its restriction to [1, 2], we can use the fact that g(2x) = g(x) for all x ∈ [1, [infinity]) and g(x) = f(x) for all x ∈ [1, 2]. Since f(x) = g(x) for x ∈ [1, 2], we can recursively use f(2x) = f(x) to show that f(x) = g(x) for all x ∈ [1, [infinity]).

3. To show that any function f satisfying these properties is uniformly continuous, we can use the fact that f(x) = f(2^n) for all x ∈ [2^n, 2^(n+1)] and n ∈ Z. Since f is continuous, it is uniformly continuous on each interval [2^n, 2^(n+1)]. By taking smaller intervals around each point, we can extend this uniform continuity to the entire domain [1, [infinity]].