6-175 Let $f$ be a function from $R \rightarrow R$, and
(1) For any $x, y \in R$,
$$f(x)+f(y)+1 \geqslant f(x+y) \geqslant f(x)+f(y)$$
(2) For any $x \in[0,1)$, $f(0) \geqslant f(x)$;
(3) $-f(-1)=f(1)=1$.

Find all functions $f$ that satisfy the conditions.
［Solution］ From (1) and (2) we get
$$2 f(0)+1 \geqslant 2 f\left(\frac{1}{2}\right)+1 \geqslant f(1) \geqslant f(1)+f(0)$$

But from (3) we know $f(1)=1$, so the above equation can be simplified to
$$2 f(0) \geqslant 0 \geqslant f(0)$$

Thus, we have $\quad f(0) \geqslant 0 \geqslant f(0)$,
$$f(0)=0 .$$

From (2), for any $x \in(0,1)$, we have
$$f(x) \leqslant 0 \text {. }$$

If there exists $x \in(0,1)$ such that $f(x)f(x)+f(1-x)+1 \geqslant f(1)=1$

it leads to a contradiction. Therefore, for any $x \in[0,1)$, we have
$$f(x)=0 \text {. }$$

From (1), for any $x \in R$, we have
$$f(x+1) \geqslant f(x)+f(1)=f(x)+1$$

Also, from (1) and (3), for any $x \in R$, we have
$$f(x) \geqslant f(x+1)+f(-1)=f(x+1)-1$$

From the above two equations, for any $x \in R$, we get
$$f(x+1)=f(x)+1$$

From (1) and (2), for any $x \in R$, we have
$$f(x)=[x]$$