# Functions That can cause Problem

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

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

for all $$x,y \in \mathbb R^{+}$$.

Note by Department 8
2 years, 7 months ago

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

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

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

And we also have that $$(f(x+f(x))=f(2x)+f(x)) \land (f(\frac{x}{2}+f(\frac{x}{2}))=f(x)+f(\frac{x}{2}) ) \implies (f(x+f(x))-f(\frac{x}{2}+f(\frac{x}{2}))=f(x)-f(\frac{x}{2}) )$$

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

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

$$f(2x+f(2x))-f(x+f(x))=f(x+f(x))-2f(x) \implies f(2x+f(2x))=2(f(x+f(x))-f(x)) \implies f(2x+f(2x))=2f(2x)$$

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

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

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

- 2 years, 7 months ago

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

- 2 years, 7 months ago