Posted by J. Holmer on September 4, 2007, 5:01 pm, in reply to "HW 1 #5" A= \sum_n \sum_m x_{m,n} B = \sum_m \sum_n x_{m,n} and then the goal is to show that A=B. Let \epsilon>0. Show (and you'll \sum_{n=1}^N \sum_{m=1}^M x_{m,n} is within \epsilon of A and \sum_{m=1}^M \sum_{n=1}^N x_{m,n} is within \epsilon of B For (b), I can't think of a good hint without basically giving the answer. For (c), you'll use part (a) and it is really a generalization of the proof JH
The rearrangement theorem is in a similar spirit but it is true that you
can't apply it directly (as far as I know). I recommend first assuming that
both orders of summing converge to some numbers:
need to use that x_{m,n} >=0 at this point) that there exists and M and N
such that
You need to use an infinite number of positive and negative numbers (but
this is probably obvious). Try to find a 2x2 array so that when you sum
over the rows, you get the totals 1/2, 1/4, 1/8, etc. and when you sum over
the columns, you get the totals -1/2, -1/4, -1/8, etc.
of the Minkowski inequality giving \|x+y\|_{\ell^p} \leq \|x\|_{\ell^p}
+\|y\|_{\ell^p}. There was one step in that proof (presented in class)
where we used that |x_j+y_j| \leq |x_j| + |y_j|, and then split the sum into
two pieces.
This is where (a) comes in.
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