sequences and series - a provable upper bound on the summation

Here is a copy-and-paste of the answer I posted to the same question on math.stackexchange.com so that the questioner can close this question.

The sum in question is at most ε². (We do not need the condition that the row sum equals 1 or the condition f(i)≠g(i) to obtain this.)

Proof. Since
$$sum_{k=1}^n(a_{f(k)j}-a_{g(k)j})=sum_{k=1}^na_{f(k)j}-sum_{k=1}^na_{g(k)j}=0,$$
we have
$$left|sum_{k=1}^na_{ki}(a_{f(k)j}-a_{g(k)j})right|
=left|sum_{k=1}^n(a_{ki}-a_{1i})(a_{f(k)j}-a_{g(k)j})right|$$
$$lesum_{k=1}^n|a_{ki}-a_{1i}||a_{f(k)j}-a_{g(k)j}|.$$
Therefore, the sum in question is at most
$$frac1nsum_{i,j=1}^zsum_{k=1}^n|a_{ki}-a_{1i}||a_{f(k)j}-a_{g(k)j}|
=frac1nsum_{k=1}^nleft(sum_{i=1}^z|a_{ki}-a_{1i}|right)left(sum_{j=1}^z|a_{f(k)j}-a_{g(k)j}|right)$$
$$lefrac1nsum_{k=1}^nepsilon^2=epsilon^2.$$