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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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.
avz2611
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From the original equation it is obvious that ax+b works... but are there any other solutions? – Fermat Feb 20 '15 at 15:04
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Hmm actually is it clear $f$ is differentiable? – Seth Feb 20 '15 at 15:06
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@Fermat no , i ensured derivative has to be constant so all polynomials that have constant derivative are included in this – avz2611 Feb 20 '15 at 15:06
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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 $f$. – Uncountable Feb 20 '15 at 15:06
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@Seth is the answer satisfactory now – avz2611 Feb 20 '15 at 15:14
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@Uncountable i proved the function has to be differentiable – avz2611 Feb 20 '15 at 15:16
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@avz2611 No, you have proven that the secants of $f$ at $x$ and $z$ are identical, so $f'(z) = f'(x)$ if either limit exists. But you haven't proved either limit exists as $y\to 0$ (they don't in general). – Erick Wong Feb 20 '15 at 15:17
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It does not prove that f is differentiable. If the limit exists then f is differentible at x...Then how do you divide both sides by y as it is not nonzero? – Fermat Feb 20 '15 at 15:18
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@avz2611 To put it another way: you have shown that the expression $[f(x+y)-f(x)]/y$ is independent of $x$ and thus may be written as $g(y)$. This doesn't say anything about whether $\lim_{y\to 0} g(y)$ exists. – Erick Wong Feb 20 '15 at 15:20
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@ErickWong if limit doesn't exist for a single point it should not exist for any point right? – avz2611 Feb 20 '15 at 15:21
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@avz2611 Correct. But many functions are nowhere-differentiable, and $f$ could very well be as such. – Erick Wong Feb 20 '15 at 15:21
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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.
John Hughes
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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. – quid Feb 20 '15 at 15:10
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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 $C$.
Ben Grossmann
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