Problem 1

Let

f:\mathbb R\to\mathbb R

satisfy

f(x)+f(y)+1\ge f(x+y)\ge f(x)+f(y)

for all real

x,y

;

f(0)\ge f(x)

for

x\in[0,1)

; and

-f(-1)=f(1)=1

. Find all such functions.

Step 4 of 4: Extend by translation and verify

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f(x)=0\ (0\le x<1),\qquad f(x)=\lfloor x\rfloor\ (x\in\mathbb R)

Detailed analysis

For $0<x<1$ , both $f(x)$ and $f(1-x)$ are nonpositive while their sum is nonnegative, so both are zero; also $f(0)=0$ . Repeatedly using $f(x+1)=f(x)+1$ gives $f(x)=\lfloor x\rfloor$ . Conversely, $0\le\{x\}+\{y\}<2$ implies $\lfloor x\rfloor+\lfloor y\rfloor\le\lfloor x+y\rfloor\le\lfloor x\rfloor+\lfloor y\rfloor+1$ , so this function satisfies all conditions.