Let f (x) be defined as (- L, l). It is proved that there must exist even function g (x) and odd function H (x) on (- L, l) such that f (x) = g (x) + H (x) Suppose g (x) and H (x) exist, such that f (x) = g (x) + H (x), (1), And G (- x) = g (x), H (- x) = - H (x) So f (- x) = g (- x) + H (- x) = g (x) - H (x), (2) Using (1) and (2), G (x) = [f (x) + F (- x)] / 2 h(x)=[f(x)-f(-x)]/2 Then G (x) + H (x) = f (x), g(-x)=[f(-x)+f(x)]/2=g(x), h(-x)=[f(-x)-f(x)]/2=h(x). I can't understand the proof process of this problem. He first assumes that G (x) and H (x) exist, and satisfies f (x) = g (x) + H (x), and obtains formula (2). Then he uses the conclusions (1) and (2) obtained from the hypothesis to prove that the original hypothetical sentence is true. How can the proof be correct, The last line should be h (- x) = [f (- x) - f (x)] / 2 = - H (x)

Let f (x) be defined as (- L, l). It is proved that there must exist even function g (x) and odd function H (x) on (- L, l) such that f (x) = g (x) + H (x) Suppose g (x) and H (x) exist, such that f (x) = g (x) + H (x), (1), And G (- x) = g (x), H (- x) = - H (x) So f (- x) = g (- x) + H (- x) = g (x) - H (x), (2) Using (1) and (2), G (x) = [f (x) + F (- x)] / 2 h(x)=[f(x)-f(-x)]/2 Then G (x) + H (x) = f (x), g(-x)=[f(-x)+f(x)]/2=g(x), h(-x)=[f(-x)-f(x)]/2=h(x). I can't understand the proof process of this problem. He first assumes that G (x) and H (x) exist, and satisfies f (x) = g (x) + H (x), and obtains formula (2). Then he uses the conclusions (1) and (2) obtained from the hypothesis to prove that the original hypothetical sentence is true. How can the proof be correct, The last line should be h (- x) = [f (- x) - f (x)] / 2 = - H (x)

The first step is to assume that the problem of proof is the condition, that is, the method of proof to the contrary
The second step can be derived from the first step
The latter is derived from the former conditions. Just compare the final results with the contradictions to be proved

Let f (x) be defined as (- L, l). It is proved that there must exist even and odd functions H (x) on (- L, l) such that f (x) = g (x) + H (x)
It is proved in the book that if G (x) and H (x) exist, f (x) = g (x) + H (x), (1),
And G (- x) = g (x), H (- x) = - H (x)
So f (- x) = g (- x) + H (- x) = g (x) - H (x), (2)
Using (1) and (2), we can make g (x) and H (x)
g(x)=[f(x)+f(-x)]/2
h(x)=[f(x)-f(-x)]/2
Then G (x) + H (x) = f (x),
g(-x)=[f(-x)+f(x)]/2=g(x),
h(-x)=[f(-x)-f(x)]/2=h(x)
My question is:
g(x)=[f(x)+f(-x)]/2
h(x)=[f(x)-f(-x)]/2
This step is obtained by assuming the existence of even function g (x) and odd function H (x). Why use it to prove the last step
g(-x)=[f(-x)+f(x)]/2=g(x),
h(-x)=[f(-x)-f(x)]/2=h(x),
Isn't it a contradiction to prove that it is a odd even function by hypothesis?

Pay attention to the sentence "this inspires us to prove it"
It has been said that the solution is to get the expressions of G (x) and H (x) by assuming even function g (x) and odd function H (x), so as to inspire us to think about how to construct g (x) and H (x)

Let f (x) be defined as D. It is proved that there must exist even function g (x) and odd function H (x) on d such that f (x) = g (x) + H (x)
So why do symmetric intervals have these two odd and even functions? How to prove it.

It is proved that such g (x) and H (x) can be constructed
Because f (x) = g (x) + H (x)
Then f (- x) = g (- x) + H (- x) = g (x) - H (x)
So g (x) = [f (x) + F (- x)] / 2, H (x) = [f (x) - f (- x)] / 2
Obviously g (x) is an even function and H (x) is an odd function