Functions and Mapping question?

Question

Determine all functions $f: \mathbb{R} \to \mathbb{R}$ satisfying the condition

$$2f(x) = f(x + y) + f(x + 2y)$$

for all real numbers $x$ and all non-negative real numbers $y$.

I just don't know where to start.

Answer 1

HINT: If you set $x=0$, the condition requires that $$2f(0)=f(y)+f(2y)$$ for all real $y\ge 0$. Thus, $$f(y)+f(2y)=2f(0)=f(2y)+f(4y)\;,$$ so $f(y)=f(4y)$ for all $y\ge 0$. This doesn’t prove that $f$ has to be a constant function, but it certainly points in that direction. This suggests that one should suppose that $f$ isn’t constant and see whether it’s possible to derive a contradiction.

Even before that one might examine a numerical example. First let’s rewrite the condition:

$$f(x)=\frac12\Big(f(x+y)+f(x+2y)\Big)\tag{1}$$

whenever $y\ge 0$. Now suppose, just to get started, that some $f(x)=2$ and $f(x+y)=1$, where $y>0$. We can use $(1)$ to find that $f(x+2y)=3$. Replacing $x$ by $x+y$ in $(1)$, we see that

$$f(x+y)=\frac12\Big(f(x+2y)+f(x+3y)\Big)\;,\tag{2}$$

and we can solve to find that $f(x+3y)=-1$. Replacing $x$ by $x+2y$ in $(1)$ gives us

$$f(x+2y)=\frac12\Big(f(x+3y)+f(x+4y)\Big)\;,\tag{3}$$

which in turn can be solved to show that $f(x+4y)=7$. But now we have a problem: replacing $y$ by $2y$ in $(1)$, we find that

$$f(x)=\frac12\Big(f(x+2y)+f(x+4y)\Big)\;,\tag{4}$$

which in our numerical example says that $2=\frac12(3+7)=5$, which is impossible.

Now generalize this example: show in similar fashion that if $f$ is not constant, then it cannot satisfy $(1)$, because then it must also satisfy $(2)$-$(4)$, and we can always derive a contradiction.

Answer 2

See my answer is x=y and simultaneously x=-y both are true but for a real number they cant be true simultaneously.

In case of mod x= mod y for x less than 0 its not possible hence we cant go for the above problem as they are fixed for a single run. So the above is false for the given mapping.