A difficulty in understanding the proof of Dini`s theorem.
$begingroup$
The question and its answer is given below:
But I am not understanding the sequence of steps in the solution what are the ideas he using, for example:
1- In the second line why he is using a subsequence ${f_{n_{k}}}$ and he did not use only apoint.
2-In the forth line why we are sure that there exists such $x_{k}$ for each $k$, and why we are doing this?
3-Also I see we arrived at a contradiction to that $f_{n}$ converges pointwisely to $f$, is my understanding correct?
Could anyone clarify these discrepancies for me please?
calculus sequences-and-series uniform-convergence
asked Jan 21 at 13:37
hopefullyhopefully
209114
$endgroup$
|
show 4 more comments
$begingroup$
The question and its answer is given below:
But I am not understanding the sequence of steps in the solution what are the ideas he using, for example:
1- In the second line why he is using a subsequence ${f_{n_{k}}}$ and he did not use only apoint.
2-In the forth line why we are sure that there exists such $x_{k}$ for each $k$, and why we are doing this?
3-Also I see we arrived at a contradiction to that $f_{n}$ converges pointwisely to $f$, is my understanding correct?
Could anyone clarify these discrepancies for me please?
calculus sequences-and-series uniform-convergence
asked Jan 21 at 13:37
hopefullyhopefully
209114
$endgroup$
1
$begingroup$
For 1, write down the negation of uniform convergence. For 2, write down the definition of a supremum. For 3, yes.
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– Mindlack
Jan 21 at 13:45
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For 1, why we are using a subsequence @Mindlack
$endgroup$
– hopefully
Jan 21 at 14:02
1
$begingroup$
Because it could happen that there is a subsequence that converges uniformly and another that does not. When you merge the two, the sequence converges pointwise but not uniformly. This won’t happen actually (which is basically inside the proof), but you don’t know that beforehand (write your definition).
$endgroup$
– Mindlack
Jan 21 at 14:07
1
$begingroup$
No, because uniform convergence for a sequence is an asymptotic property. You can modify any finite number of terms of the sequence and it does not change whether uniform convergence occurs.
$endgroup$
– Mindlack
Jan 21 at 14:18
1
$begingroup$
Uniform convergence is: $forall epsilon >0,,exists N>0,,forall k geq N,, forall x,, |f_k(x)-f(x)| leq epsilon$. So if you have just one term such that $|f_k(x)-f(x)| > epsilon$ for some $x$, it does not violate the property... you just choose $N > k$.
$endgroup$
– Mindlack
Jan 21 at 14:28
|
show 4 more comments
$begingroup$
The question and its answer is given below:
But I am not understanding the sequence of steps in the solution what are the ideas he using, for example:
1- In the second line why he is using a subsequence ${f_{n_{k}}}$ and he did not use only apoint.
2-In the forth line why we are sure that there exists such $x_{k}$ for each $k$, and why we are doing this?
3-Also I see we arrived at a contradiction to that $f_{n}$ converges pointwisely to $f$, is my understanding correct?
Could anyone clarify these discrepancies for me please?
calculus sequences-and-series uniform-convergence
asked Jan 21 at 13:37
hopefullyhopefully
209114
$endgroup$
The question and its answer is given below:
But I am not understanding the sequence of steps in the solution what are the ideas he using, for example:
1- In the second line why he is using a subsequence ${f_{n_{k}}}$ and he did not use only apoint.
2-In the forth line why we are sure that there exists such $x_{k}$ for each $k$, and why we are doing this?
3-Also I see we arrived at a contradiction to that $f_{n}$ converges pointwisely to $f$, is my understanding correct?
Could anyone clarify these discrepancies for me please?
calculus sequences-and-series uniform-convergence
calculus sequences-and-series uniform-convergence
asked Jan 21 at 13:37
hopefullyhopefully
209114
asked Jan 21 at 13:37
hopefullyhopefully
209114
asked Jan 21 at 13:37
hopefullyhopefully
209114
asked Jan 21 at 13:37
hopefullyhopefully
209114
asked Jan 21 at 13:37
hopefullyhopefully
209114
209114
1
$begingroup$
For 1, write down the negation of uniform convergence. For 2, write down the definition of a supremum. For 3, yes.
$endgroup$
– Mindlack
Jan 21 at 13:45
$begingroup$
For 1, why we are using a subsequence @Mindlack
$endgroup$
– hopefully
Jan 21 at 14:02
1
$begingroup$
Because it could happen that there is a subsequence that converges uniformly and another that does not. When you merge the two, the sequence converges pointwise but not uniformly. This won’t happen actually (which is basically inside the proof), but you don’t know that beforehand (write your definition).
$endgroup$
– Mindlack
Jan 21 at 14:07
1
$begingroup$
No, because uniform convergence for a sequence is an asymptotic property. You can modify any finite number of terms of the sequence and it does not change whether uniform convergence occurs.
$endgroup$
– Mindlack
Jan 21 at 14:18
1
$begingroup$
Uniform convergence is: $forall epsilon >0,,exists N>0,,forall k geq N,, forall x,, |f_k(x)-f(x)| leq epsilon$. So if you have just one term such that $|f_k(x)-f(x)| > epsilon$ for some $x$, it does not violate the property... you just choose $N > k$.
$endgroup$
– Mindlack
Jan 21 at 14:28
|
show 4 more comments
1
$begingroup$
For 1, write down the negation of uniform convergence. For 2, write down the definition of a supremum. For 3, yes.
$endgroup$
– Mindlack
Jan 21 at 13:45
$begingroup$
For 1, why we are using a subsequence @Mindlack
$endgroup$
– hopefully
Jan 21 at 14:02
1
$begingroup$
Because it could happen that there is a subsequence that converges uniformly and another that does not. When you merge the two, the sequence converges pointwise but not uniformly. This won’t happen actually (which is basically inside the proof), but you don’t know that beforehand (write your definition).
$endgroup$
– Mindlack
Jan 21 at 14:07
1
$begingroup$
No, because uniform convergence for a sequence is an asymptotic property. You can modify any finite number of terms of the sequence and it does not change whether uniform convergence occurs.
$endgroup$
– Mindlack
Jan 21 at 14:18
1
$begingroup$
Uniform convergence is: $forall epsilon >0,,exists N>0,,forall k geq N,, forall x,, |f_k(x)-f(x)| leq epsilon$. So if you have just one term such that $|f_k(x)-f(x)| > epsilon$ for some $x$, it does not violate the property... you just choose $N > k$.
$endgroup$
– Mindlack
Jan 21 at 14:28
1
1
$begingroup$
For 1, write down the negation of uniform convergence. For 2, write down the definition of a supremum. For 3, yes.
$endgroup$
– Mindlack
Jan 21 at 13:45
$begingroup$
For 1, write down the negation of uniform convergence. For 2, write down the definition of a supremum. For 3, yes.
$endgroup$
– Mindlack
Jan 21 at 13:45
$begingroup$
For 1, why we are using a subsequence @Mindlack
$endgroup$
– hopefully
Jan 21 at 14:02
$begingroup$
For 1, why we are using a subsequence @Mindlack
$endgroup$
– hopefully
Jan 21 at 14:02
1
1
$begingroup$
Because it could happen that there is a subsequence that converges uniformly and another that does not. When you merge the two, the sequence converges pointwise but not uniformly. This won’t happen actually (which is basically inside the proof), but you don’t know that beforehand (write your definition).
$endgroup$
– Mindlack
Jan 21 at 14:07
$begingroup$
Because it could happen that there is a subsequence that converges uniformly and another that does not. When you merge the two, the sequence converges pointwise but not uniformly. This won’t happen actually (which is basically inside the proof), but you don’t know that beforehand (write your definition).
$endgroup$
– Mindlack
Jan 21 at 14:07
1
1
$begingroup$
No, because uniform convergence for a sequence is an asymptotic property. You can modify any finite number of terms of the sequence and it does not change whether uniform convergence occurs.
$endgroup$
– Mindlack
Jan 21 at 14:18
$begingroup$
No, because uniform convergence for a sequence is an asymptotic property. You can modify any finite number of terms of the sequence and it does not change whether uniform convergence occurs.
$endgroup$
– Mindlack
Jan 21 at 14:18
1
1
$begingroup$
Uniform convergence is: $forall epsilon >0,,exists N>0,,forall k geq N,, forall x,, |f_k(x)-f(x)| leq epsilon$. So if you have just one term such that $|f_k(x)-f(x)| > epsilon$ for some $x$, it does not violate the property... you just choose $N > k$.
$endgroup$
– Mindlack
Jan 21 at 14:28
$begingroup$
Uniform convergence is: $forall epsilon >0,,exists N>0,,forall k geq N,, forall x,, |f_k(x)-f(x)| leq epsilon$. So if you have just one term such that $|f_k(x)-f(x)| > epsilon$ for some $x$, it does not violate the property... you just choose $N > k$.
$endgroup$
– Mindlack
Jan 21 at 14:28
|
show 4 more comments
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1
$begingroup$
For 1, write down the negation of uniform convergence. For 2, write down the definition of a supremum. For 3, yes.
$endgroup$
– Mindlack
Jan 21 at 13:45
$begingroup$
For 1, why we are using a subsequence @Mindlack
$endgroup$
– hopefully
Jan 21 at 14:02
1
$begingroup$
Because it could happen that there is a subsequence that converges uniformly and another that does not. When you merge the two, the sequence converges pointwise but not uniformly. This won’t happen actually (which is basically inside the proof), but you don’t know that beforehand (write your definition).
$endgroup$
– Mindlack
Jan 21 at 14:07
1
$begingroup$
No, because uniform convergence for a sequence is an asymptotic property. You can modify any finite number of terms of the sequence and it does not change whether uniform convergence occurs.
$endgroup$
– Mindlack
Jan 21 at 14:18
1
$begingroup$
Uniform convergence is: $forall epsilon >0,,exists N>0,,forall k geq N,, forall x,, |f_k(x)-f(x)| leq epsilon$. So if you have just one term such that $|f_k(x)-f(x)| > epsilon$ for some $x$, it does not violate the property... you just choose $N > k$.
$endgroup$
– Mindlack
Jan 21 at 14:28