If we define $x, y \in \mathbb{R}$, is there a function that fulfills the condition:
$f(x+y) = f(xy)$
for all $x$ and $y$?
For now, let's assume there are no stipulations on continuity and differentiability.
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1Any constant function works. – Adrian Keister Aug 29 '19 at 19:24
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Constant functions will satisfy this property. – Aniruddha Deshmukh Aug 29 '19 at 19:24
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2Taking $y=0$ we see that $f(x)=f(0)$ identically. – lulu Aug 29 '19 at 19:25
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Choosing $y=0$, we have $f(x) = f(0)$ for all $x$ – amsmath Aug 29 '19 at 19:25
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3Only constant functions would work, because $f(x+0)=f(0)$ for all $x$ – Hyperion Aug 29 '19 at 19:25
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Would anything beyond a constant function work? I should have excluded that case. – dsmalenb Aug 29 '19 at 19:26
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I also specified for all x and y in R. These answers do not work for that condition. – dsmalenb Aug 29 '19 at 19:27
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5Have you read the comments? Several people have shown that only constants work. – lulu Aug 29 '19 at 19:28
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@dsmalenb: Just because the function is constant doesn't mean that it doesn't work for all $x,y\in\mathbb{R}.$ Let $f(x)=\text{const};\forall,x\in\mathbb{R}.$ This function satisfies all constraints, and as a number of folks have shown in the comments, this is the only kind of solution. – Adrian Keister Aug 29 '19 at 19:30
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@projectilemotion, you are correct. – NoChance Aug 29 '19 at 19:42
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@dsmalenb I believe the issue is you are concerned that a constant function will not satisfy the conditions, and those who are commenting are glossing over the fact that they do satisfy the conditions. Suppose $f(x)=c$ is a constant function. Then $f(x+y) = f(xy) = c$ for all $x,y \in \mathbb{R}$. Is this the proof your are looking for? – SlipEternal Aug 29 '19 at 20:30
1 Answers
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If $y=0$ then $f(x)=f(x+0)=f(0x)=f(0)$, so $f$ is a constant.
dcolazin
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Is your response answering the OP question? OP says: "is there a function that fulfills the condition:...."? – NoChance Aug 29 '19 at 20:15
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@NoChance The title question hardly needs an answer, because $f=0$ is certainly such a function. So it makes sense to give an answer which takes into account the comments, too. – Dietrich Burde Aug 29 '19 at 20:24
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The problem here is that the domain of $f$ is not correctly specified in the question. – IV_ Aug 29 '19 at 20:42