I can not understand a sentence in a proof, can you help me ?
water
Now I write the theorem and its proof.
Theorem: Let f_1,f_2,...,f_n,....are increasing functions in [a,b].
f(x) = sum_1^oo f_n (x), x in [a,b].
Then f is increasing function and f ' (x) = sum_1^oo f_n'(x) a.e. in
[a,b].
Here "'" is the first derivative, a.e. is almost everywhere.
Proof: WLOG we assume f_n(a) = 0, n=1,2,.....
Let S_n (x) = sum_1^n f_i (x).
We can easily find that S_n (x) and f (x) are increasing functions in
[a,b].
So there exists set E in [a,b], m (E) = 0,
For any x in [a,b] - E, f_1'(x),...,f_n'(x),...,f'(x) all exist and are
finite.
Denote S_0(x)=0, S_n (x) - S_{n-1} (x) = f_n (x) and f (x) - S_n (x) =
sum_{n+1}^oo f_i (x) are increasing functions.
When x in [a,b] - E, [ S_n (x) - S_{n-1} (x) ] ' >= 0, [ f (x) - S_n
(x) ] ' >= 0.
So S_{n-1} ' (x) <= S_n ' (x) <= f ' (x), x in [a,b] - E.
For any x in [a,b] - E, lim_{n->oo} S_n' exists and are finite.
Sum_1^oo f_n'(x) = lim_{n->oo} sum_1^n f_i ' (x) = lim_{n->oo}
S_n'(x), x in [a,b] - E.
Now we shall prove there exists subsequence S_{n_k}'(x) of S_n'(x),
S_{n_k}'(x) -> f"(x) a.e.[a,b]. So f '(x) = sum_1^oo f_n'(x) a.e.
[a,b].
Because lim_{n->oo} S_n(b)=f(b), so for any k in N, exists n_k in N ,
f(b) - S_{n_k}(b) < 1/(2^k).
Because f(x) - S_{n_k}(x) is increasing and f(a) - S_{n_k}(a)=0,
0 <= sum_1^oo [f(x) - S_{n_k}(x)] <= sum_1^oo [f(b) - S_{n_k}(b)] <
sum_1^oo 1/(2^k) = 1.
Until now I am clear. But the following sentence makes me confused.
According above fact, we have sum_1^oo [f'(x) - S_{n_k} ' (x)] < +oo
a.e. [a,b].
Why?
From the convergence of function series, we conclude S_{n_k}'(x) ->
f'(x) a.e. [a,b].
Can you explain to me the sentense "According above fact, we have
sum_1^oo [f'(x) - S_{n_k}'(x)] < +oo a.e. [a,b]"?
Thanks.
water
Sorry , here S_{n_k}'(x) -> f '(x) a.e.[a,b]
A friendly output can be seen http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1021334#1021334
water
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Can you help me? I am nervous.
water
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can you give me some clues?
water
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waiting on line.
Can you help me ?
Rotwang
I think perhaps this is supposed to follow from an application of the
fundamental theorem of calculus together with the monotone convergence
theorem. Note that, from the former, we have that
[f(b) - S_{n_k}(b)] = [f(b) - S_{n_k}(b)] - [f(a) - S_{n_k}(a)] =
Int_a^b[f'(x) - S_{n'_k}(x)] dx
provided that b is in E - otherwise, since all f_n are increasing, we
will have that the LHS is >= the RHS (I am not certain about this
step, since I'm not sufficiently familiar with this stuff to know the
precise statement of the theorem by heart and I don't have an
appropriate book to hand - I will check later). Then use the monotone
convergence theorem to equate the sum of terms on the right hand side
to the integral of sum_1^oo [f'(x) - S_{n_k}'(x)]; this must then be
finite almost everywhere if its integral is finite.
Rotwang
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> I think perhaps this is supposed to follow from an application of the
> fundamental theorem of calculus together with the monotone convergence
> theorem. Note that, from the former, we have that
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> [f(b) - S_{n_k}(b)] = [f(b) - S_{n_k}(b)] - [f(a) - S_{n_k}(a)] =
> Int_a^b[f'(x) - S_{n'_k}(x)] dx
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> provided that b is in E
Sorry, this is obviously bullshit. But what isn't bullshit is Theorem
2 of chapter 5 of Royden's "Real Analysis", which states that, given
an increasing real-valued function F on [a,b], F is differentiable
almost everywhere, the derivative F' is measurable, and
Int_a^b F'(x) dx <= F(b) - F(a)
Applying this to the above functions we find that
[f(b) - S_{n_k}(b)] >= Int_a^b[f'(x) - S_{n'_k}(x)] dx
The rest of what I wrote was fine I think.
water
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Thanks very much! You help me.