Find $f(x)$, given $f(x+y) - 2f(x-y) + f(x) - 2f(y) = y - 2$

Question

For a differentiable function $f(x)$, $$f(x+y) - 2f(x-y) + f(x) - 2f(y) = y - 2.$$ Then find $f(x)$.

I am not able to think of any method which is working. I tried replacing $x$ and $y$, putting $x$ or $y = 0$, putting $y = x$, etc. Can someone help me

Answer given at back is $f(x)=x+1$

Answer 1

Another trick for solving. As PNDas has already mentioned $f(0) = 1$. Now substitute $y = h$ and do some grouping:

$$(f(x+h) - f(x)) - 2(f(x +(-h)) - f(x)) = h + 2(f(0 + h) - f(0))$$

Divide by $h$

$$\frac{f(x+h) - f(x)}h +2\frac{f(x +(-h)) - f(x)}{-h} = 1 +2\frac{f(0 + h) - f(0)}h$$ And take the limit as $h \to 0$ to get $$3f'(x) = 1 + 2f'(0)$$ which by the assumption of differentiability is guaranteed to converge. Set $x = 0$ to find $f'(0)$, then it is a simple integration to solve.

Answer 2

You can show that the only function $ f : \mathbb R \to \mathbb R $ satisfying $$ f ( x + y ) - 2 f ( x - y ) + f ( x ) - 2 f ( y ) = y - 2 \tag 0 \label 0 $$ for all $ x , y \in \mathbb R $ is $ f ( x ) = x + 1 $, even without assuming that $ f $ is differentiable. It's easy to check that the mentioned function is a solution. To prove the converse, first define $ g : \mathbb R \to \mathbb R $ with $ g ( x ) = f ( x ) - x - 1 $. Then \eqref{0} can easily be rephrased in terms of $ g $ to get $$ g ( x + y ) - 2 g ( x - y ) + g ( x ) - 2 g ( y ) = 0 \text . \tag 1 \label 1 $$ Letting $ x = y = 0 $ in \eqref{1} we get $ g ( 0 ) = 0 $. Now, letting $ x = 0 $ in \eqref{1} we get $$ - 2 g ( - y ) = g ( y ) \text . \tag 2 \label 2 $$ Substituting $ - y $ for $ y $ in \eqref{2} we get $ g ( - y ) = - 2 g ( y ) $, which together with \eqref{2} itself shows that $ g ( y ) = 0 $, and thus $ f ( y ) = y + 1 $ for all $ y \in \mathbb R $.

Answer 3

Solution based on the one provided by user FhoToN in this AoPS post. (no differentiability needed nor used)

Let $P(x,y)$ be the assertion that $$f(x+y)-2f(x-y)+f(x)-2f(y)=y-2.$$

From $P(x,0)$ we get $-2f(0)=-2$, and so $$f(0)=1.$$

From $P(0,y)$ and previous we get $$f(y)+2f(-y)=3-y.\tag{1}$$

From $P(0,-y)$ and previous we get $$f(-y)+2f(y)=3+y.\tag{2}$$

Solving $(1)$ and $(2)$ as a system of equations (with variables $f(y)$ and $f(-y)$), we get $f(y)=y+1$ (notice that $y$ was arbitrary chosen and so this holds for all $y$). We can verify this is a solution, and thus the only one.