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I have $f(x,y):R^2\rightarrow R$. I know $f(x,y)=f(y,x)$ and $f(x+d,y)=f(x,y+d)$. Can I prove that I can express $f(x,y)$ as $g(x+y)$.

This is where I got: $f(x+d,y)=f(x,y+d)$, I plug in $x=0$

Gives me $f(d,y)=f(0,y+d)$, I plug in $d=x$, and use the symmetry condition

this gives $f(x,y)=f(x+y,0)$.

I have a strong hunch that this means that $f(x,y)$ can be expressed as $g(x+y)$, but I am not sure how to justify my hunch. Can you help me?

Tuomas Lahtinen
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1 Answers1

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As you pointed out, for any $d$, $ f(x,y) = f(x-d,y+d) $.

With $d = x$ and $g(t)=f(0,t)$

$$ f(x,y)=f(0,x+y)=g(x+y). $$

Note that you don't need to use the symmetry property, which is a trivial consequence $f(y,x)=g(y+x)=f(x,y)$.

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