6-139 Find a function $f(m, n)$ that satisfies the following conditions: for every pair of non-negative integers $m, n$,
(1) $2 f(m, n)=2+f(m+1, n-1)+f(m-1, n+1)$,
$(m \geqslant 1, n \geqslant 1)$
(2) $f(m, 0)=f(0, n)=0$.
[Solution] From (1)
$$
\begin{aligned}
& f(m, n)-f(m-1, n+1) \\
= & f(m+1, n-1)-f(m, n)+2,
\end{aligned}
$$

In the above equation, replace $m, n$ with $m-k, n+k$ (non-negative integer $k \leqslant m$), we get
$$
\begin{aligned}
& f(m-k, n+k)-f(m-k-1, n+k+1) \\
= & f(m-k+1, n+k-1)-f(m-k, n+k)+2,
\end{aligned}
$$

Substitute $k=0,1,2, \cdots, m-1$ into the above equation, we get $m$ equations. Adding these $m$ equations, we have
$$
\begin{aligned}
& \sum_{K=0}^{m-1}\{f(m-k, n+k)-f(m-k-1, n+k+1)\} \\
= & \sum_{K=0}^{m-1}\{f(m-k+1, n+k-1)-f(m-k, n+k)+2\},
\end{aligned}
$$

Using (2) and simplifying, we get
$$
f(m, n)=f(m+1, n-1)-f(1, n+m-1)+2 m,
$$

which is
$$
f(m+1, n-1)-f(m, n)=f(1, n+m-1)-2 m,
$$

In the above equation, replace $m, n$ with $m+k, n-k$ (non-negative integer $k \leqslant n$), we get
$$
\begin{aligned}
& f(m+k+1, n-k-1)-f(m+k, n-k) \\
= & f(1, n+m-1)-2(m+k),
\end{aligned}
$$

Substitute $k=0,1,2, \cdots, n-1$ into the above equation, we get $n$ equations. Adding these $n$ equations, we have
$$
\begin{aligned}
& \sum_{K=0}^{n-1}\{f(m+k+1, n-k-1)-f(m+k, n-k)\} \\
= & n f(1, n+m-1)-2 \sum_{K=0}^{n-1}(m+k),
\end{aligned}
$$

Using (2) and simplifying, we get
$$
-f(m, n)=n f(1, n+m-1)-2 m n-n(n-1),
$$

which is
$$
f(m, n)=(2 m+n-1) n-n f(1, n+m-1) .
$$

In (1), replace $m, n$ with $m-k, n+k$ (non-negative integer $k \leqslant m$), we get
$$
\begin{aligned}
& f(m-k+1, n+k-1)-f(m-k, n+k) \\
= & f(1, n+m-1)-2(m-k),
\end{aligned}
$$

Substitute $k=1,2, \cdots, m$ into the above equation, we get $m$ equations. Adding these $m$ equations, we have
$$
\begin{aligned}
& \sum_{K=1}^{m}\{f(m-k+1, n+k-1)-f(m-k, n+k)\} \\
= & m f(1, n+m-1)-2 \sum_{K=1}^{m}(m-k),
\end{aligned}
$$

Using (2) and simplifying, we get
$$
f(m, n)=m f(1, n+m-1)-m(m-1)
$$

Multiplying (2) by $m$ and (3) by $n$, we get
$$
(m+n) f(m, n)=m n(m+n),
$$

Therefore, when $m, n$ are not both zero, we have
$$
f(m, n)=m n
$$

From (2), we know that the above equation also holds for $m=0$ and $n=0$.
Upon verification, the function $f(m, n)=m n$ indeed satisfies all the conditions given in the problem.
Thus, the required function is $f(m, n)=m n$.