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Is it true that $$f_{n-1}=g_n , n = 1,..,N \iff f_{n-1}+g_{n+1}=f_n + g_n , n = 1,..,N-1 $$ ? where $f,g\in[0,1]$ .

It's easy to see the sufficiency , this may be elementary but I've a hard time showing the necessity . I tried swaping them then suming them , but that's only telescoping which doen't give the LHS .

You may ignore this but this question actually came from here where user Did claimed that $*$ and $\dagger$ are equivalent , Did 's been offline for 3 yrs so I give up asking Did .

Bill Dubuque
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    Suppose $f_n = 1$ for all $n$ and $g_n = 0$ for all $n$. Here we would have $f_{n-1}+g_{n+1}=1+0$ is indeed equal to $f_n + g_n=1+0$ however $f_{n-1}\neq g_n$ as $1\neq 0$ – JMoravitz Jun 27 '22 at 13:42
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    Yeah, the best you can get is that $f_{n-1}-g_n$ is constant iff $\dots,$ which is obvious since the right hand can be written as $g_{n+1}-f_n=g_n-f_{n-1}.$ – Thomas Andrews Jun 27 '22 at 13:46

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The necessity is false , from counterexample by JMoravitz in the comment . Nonetheless , the RHS implies the LHS with the additional condition that $\exists k\in \{1,..,N-1\} , f_{k-1} = g_k $ .

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