Functions That can cause Problem

Find all such functions f:R+→R+f: \mathbb{ R }^{ + }\rightarrow \mathbb{ R }^{ + } such that

f(x+f(y))=f(x+y)+f(y)\large{f\left( x+f\left( y \right) \right) =f\left( x+y \right) +f\left( y \right) }

for all x,y∈R+x,y \in \mathbb R^{+}.

#Algebra #Functions

Note by Department 8
5 years, 7 months ago

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Comments

@Sharky Kesa , @Nihar Mahajan ,@Satyajit Mohanty ,@Sandeep Bhardwaj ,@Otto Bretscher please share this and invite other people to solve this question.

Department 8 - 5 years, 7 months ago

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@Chew-Seong Cheong , @Calvin Lin ,

Department 8 - 5 years, 7 months ago

Let f∈F(R+,R+)f \in \mathcal{F}(\mathbb{R}^+,\mathbb{R}^+) such that ∀x,y∈R+,f(x+f(y))=f(x+y)+f(y) \forall x,y \in \mathbb{R}^+, f(x+f(y))=f(x+y)+f(y) .

Let x∈R+x \in \mathbb{R}^+ . We have that

f(x+f(x))=f(2x)+f(x)f(x+f(x))=f(2x)+f(x) (1.)

And we also have that (f(x+f(x))=f(2x)+f(x))∧(f(x2+f(x2))=f(x)+f(x2))  ⟹  (f(x+f(x))−f(x2+f(x2))=f(x)−f(x2))(f(x+f(x))=f(2x)+f(x)) \land (f(\frac{x}{2}+f(\frac{x}{2}))=f(x)+f(\frac{x}{2}) ) \implies (f(x+f(x))-f(\frac{x}{2}+f(\frac{x}{2}))=f(x)-f(\frac{x}{2}) )

So, ∀x∈R+,f(2x+f(2x))−f(x+f(x))=f(2x)−f(x) \forall x \in \mathbb{R}^+, f(2x+f(2x))-f(x+f(x))=f(2x)-f(x)

Let x∈R+x \in \mathbb{R}^+ . We have, by (1.), that

f(2x+f(2x))−f(x+f(x))=f(x+f(x))−2f(x)  ⟹  f(2x+f(2x))=2(f(x+f(x))−f(x))  ⟹  f(2x+f(2x))=2f(2x) f(2x+f(2x))-f(x+f(x))=f(x+f(x))-2f(x) \implies f(2x+f(2x))=2(f(x+f(x))-f(x)) \implies f(2x+f(2x))=2f(2x)

So, ∀x∈R+,f(x+f(x))=2f(x) \forall x \in \mathbb{R}^+, f(x+f(x))=2f(x). Let x∈R+x \in \mathbb{R}^+ . In these conditions we have that

f(x+f(x))=2f(x))  ⟹  f(2x)+f(x)=2f(x)  ⟹  f(2x)=f(x) f(x+f(x))=2f(x)) \implies f(2x)+f(x)=2f(x) \implies f(2x)=f(x) (2.)

It is easy to verify that (2.) implies that ∃c∈R:f=c \exists c \in \mathbb{R}: f=c and it is also easy to verify that f=0f=0 is the particular solution.

Paulo Guilherme Santos - 5 years, 6 months ago
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