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Determine all functions $f : \mathbb{R} \to \mathbb{R}$ that satisfy the functional equation $$f(x + y) + f(z) = f(x) + f(y + z)$$ for all $x, y, z \in \mathbb{R}$.

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I claim that $f(x)$ is a solution if and only if it is of the form $c+ h(x)$ where $h(x)$ is additive that is $h(x+y)= h(x)+h(y)$.

To check that $c+ h(x)$ is a solution. is a direct calculation, To see the converse note that $f(\cdot)-f(0)$ is additive (setting $x=0$ and using the definition).

Additive functions are linear under mild regularity assumptions, especially if they are contiuous, yet not in general. For what one can say in general about additve functions see Overview of basic facts about Cauchy functional equation as a start.

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Let $$ g(x) = f(x) - f(0), so \\ f(x) = g(x) + g(0). $$ Then the defining equation reads $$ f(x + y) + f(z) = f(x) + f(y + z)\\ g(x+y) + g(0) + g(z) + f(0) = f(x) + f(0) + f(y+z) + f(0) \\ g(x+y) + g(z) = g(x) + g(y+z). $$ In other words, $g$ satisfies the same equation as $f$. But $g(0) = f(0) - f(0) = 0$.

Now look at the definining equation (for $g$) for the case $z = 0$: we get $$ g(x+y) + g(0) = g(x) + g(y) \\ g(x+y) + 0 = g(x) + g(y) \\ g(x+y) = g(x) + g(y) $$ Therefore $g$ is additive on the reals. Does this actually make it linear? I suspect so, but lack a proof.

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  • $\begingroup$ An additive function is not necessarily linear. One gets it is $\mathbb{Q}$-lineer but when you have a basis of the reals a s a rational vectorspace you can choose whatever you want for each element of the basis. $\endgroup$ – quid Feb 20 '15 at 15:10
  • $\begingroup$ en.wikipedia.org/wiki/Cauchy%27s_functional_equation $\endgroup$ – Erick Wong Feb 20 '15 at 15:13
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using partial differential on $x$ i can say $$f'(x)=f'(x+y)$$ as y can be varied the derivative of the function has to be constant. so all function of form $ax+b$ will work. as seth pointed out it is not clear function is differentiable we can take another approach. now $$f(x+y)+f(y)=f(x)+f(y+z)$$ $$f(x+y)-f(x)=f(z+y)-f(y)$$ $$\frac{f(x+y)-f(x)}{y}=\frac{f(z+y)-f(y)}{y}$$ as $y$ tends to zero $f'(x)=f'(z)$ as $x$ and $z$ are independent function is differentiable and has a constant derivative.

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  • $\begingroup$ From the original equation it is obvious that ax+b works... but are there any other solutions? $\endgroup$ – Fermat Feb 20 '15 at 15:04
  • $\begingroup$ Hmm actually is it clear $f$ is differentiable? $\endgroup$ – Seth Feb 20 '15 at 15:06
  • $\begingroup$ @Fermat no , i ensured derivative has to be constant so all polynomials that have constant derivative are included in this $\endgroup$ – avz2611 Feb 20 '15 at 15:06
  • $\begingroup$ If $f$ is differentiable then these are the only solutions, since (for completeness) if $f(x)=ax+b$ then: $$f(x+y)+f(z)=a(x+y)+b+az+b=ax+b+a(y+z)+b=f(x)+f(y+z)$$ Also as Seth says we still have to consider nondifferentiable $f$. $\endgroup$ – Uncountable Feb 20 '15 at 15:06
  • $\begingroup$ @Seth is the answer satisfactory now $\endgroup$ – avz2611 Feb 20 '15 at 15:14
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We can rewrite the equation as $$ f(y+x)-f(x) = f(y+z) - f(z) $$ for all $x,y,z$. This is equivalent to the statement $$ f(y+x) - f(x) = C(y) $$ for some function $C: \Bbb R \to \Bbb R$. It follows that $$ f(y) - f(0) = C(y) \implies f(y) = f(0) + C(y)\\ f(0) - f(0) = C(0) \implies C(0) = 0 $$ Define $D = f(0)$. The equation we started with is equivalent to $f(x) = C(x) + D$ where $C(x)$ and $D$ are such that $$ C(y+x) - C(x) = C(y) \\ f(0) = D $$ That is, $f$ is the sum of a constant $D$ and any additive function $C$.

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