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Let $f$ be a continuous real-to-real function that satisfies $f(0)=1$ and $$f(m+n+1)=f(m)+f(n)$$ for all $m,n\in{\mathbb R}$. Show $f (x) = x + 1$ for all $x$.

What is the solution?

Source: Solving mathematical problems a personal perspective, exercise 3.1

MENZIES
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  • What is the question and where is your work? Why do keep writing $m$ and $n$ and then say for all $x$? Please edit the question carefully so that we can understand it. – Kavi Rama Murthy Oct 05 '19 at 05:32

2 Answers2

1

Suppuse $f$ is continuous with $f(0)=1$ and satisfies for each $x$, $y$: $$f(x+y+1)=f(x)+f(y)\tag1$$ Then $f(x)=f(x-1+0+1)\stackrel{(1)}=f(x-1)+f(0)=f(x-1)+1$, so $$f(x-1)=f(x)-1.\tag2$$ Define $$g(x):=f(x)-1.\tag3$$ Then for all $x$, $y$ we have $$\begin{align} g(x+y)\stackrel{(3)} =&f(x+y)-1\\ =&f(x+y-1+1)-1\\ \stackrel{(1)}=& f(x)+f(y-1)-1\\ \stackrel{(2)}=&f(x)+f(y)-1-1\\ \stackrel{(3)}=&g(x)+g(y). \end{align}$$ We've shown that $g(x+y)=g(x)+g(y)$. So $g$ satisfies the Cauchy functional equation. Since $g$ is continuous, conclude $g(x)=cx$ for some constant $c$. Plug this back into (3) to find $f(x)=cx+1$. The only value of $c$ that allows $f$ to satisfy (2) is $c=1$ and therefore $f(x)=x+1$. Check that this choice of $f$ satisfies (1).

grand_chat
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0

My solution:

$f(m+n+1+1)$

$=f(m+1)+f(n)$

$=f(m)+f(n)+f(0)$

$=f(m+n+1)+1$

Then,

$f(x+1)=f(x)+1$

So,

$f(x)=x+1$

MENZIES
  • 135
  • Are all the criteria of the desired function met with this choice in f? If so, then you've done everything the problem asked. – infinitylord Oct 05 '19 at 05:38
  • If x is an integer, my choice in f will meet the criteria. But if x is a rational number or a real number, I don't find how to show f meet the criteria. – MENZIES Oct 05 '19 at 06:01