MathLabs

Problem 1

Let f:R→Rf:\mathbb R\to\mathbb R satisfy f(x)+f(y)+1≥f(x+y)≥f(x)+f(y)f(x)+f(y)+1\ge f(x+y)\ge f(x)+f(y) for all real x,yx,y; f(0)≥f(x)f(0)\ge f(x) for x∈[0,1)x\in[0,1); and −f(−1)=f(1)=1-f(-1)=f(1)=1. Find all such functions.
Step 2 of 4: Apply it with y=−1y=-1
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f(x)≤f(x+1)≤f(x)+1f(x)\le f(x+1)\le f(x)+1
Detailed analysis

Put y=−1y=-1 and replace xx by x+1x+1. Since f(−1)=−1f(-1)=-1, the inequality becomes f(x)≤f(x+1)≤f(x)+1f(x)\le f(x+1)\le f(x)+1. Combining with the previous bounds yields f(x+1)=f(x)+1f(x+1)=f(x)+1.