6 Let $\alpha$ be a given positive real number, find all functions $f: \mathbf{N}^{*} \rightarrow \mathbf{R}$, such that for any positive integers $k, m$ satisfying $\alpha m \leqslant k<(\alpha+1) m$, we have
$$
f(k+m)=f(k)+f(m) .
$$
(Chen Yonggao)
Solution: $f(n)=b n, b$ is any given real number.
First, we prove: when $n \geqslant 2 \alpha+3$, we have
$$
f(n+1)-f(n)=f(n)-f(n-1) .
$$

This is to prove that there exists a positive integer $u$ such that
$$
\begin{array}{l}
f(n+1)-f(n)=f(u+1)-f(u), \\
f(n)-f(n-1)=f(u+1)-f(u) .
\end{array}
$$

Where (1) is equivalent to
$$
f(n+1)+f(u)=f(n)+f(u+1) .
$$

By the given condition, this is to have a positive integer $u$ satisfying
$$
\alpha(n+1) \leqslant u1$ when $ \alpha > 1$, take a positive integer $k$ such that $\alpha n_{1} \leqslant kn_{1}, k+n_{1}>n_{1}, f\left(k+n_{1}\right)=f(k)+f\left(n_{1}\right) .
$$

Therefore,
$$
f\left(n_{1}\right)=f\left(k+n_{1}\right)-f(k)=\left(k+n_{1}\right) b-k b=n_{1} b,
$$

which is a contradiction.
When $\alpha \leqslant 1$, we have $\alpha n_{1} \leqslant n_{1}<(\alpha+1) n_{1}$. Therefore,
$$
b \cdot 2 n_{1}=f\left(2 n_{1}\right)=f\left(n_{1}\right)+f\left(n_{1}\right),
$$

thus $f\left(n_{1}\right)=n_{1} b$, which is a contradiction.
Finally, for any real number $b$, the function $f(n)=b n$ clearly meets the requirements of the problem, which proves our assertion.