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 have
$$
2 f(0)+1 \geqslant 2 f\left(\frac{1}{2}\right)+1 \geqslant f(1) \geqslant f(1)+f(0),
$$

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

Thus,
$$
\begin{array}{l}
f(0) \geqslant 0 \geqslant f(0), \\
f(0)=0 .
\end{array}
$$

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

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 inequalities, for any $x \in R$, we have
$$
f(x+1)=f(x)+1 \text {. }
$$

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