Mixing
For metric space $C^0([0,1])$, a sequence of functions ${x^n}$ is Cauchy in $||-||_1$ norm but not in...
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In my text book, it is stated that for a sequence of functions ${x^n}$ in a metric space $C^0([0,1])$, it is Cauchy in $||-||_1$ norm but not in $||-||_infty$ norm because that would lead to a contradiction.
I am wondering what this contradiction is.
I tried expanding the norm:
$||x^m - x^n||_1 = int_0^1|x^m-x^n| dx =_{| m<n} int_0^1(x^m-x^n) dx = frac{1}{m+1} - frac{1}{n+1}$, which goes to zero (which is in C) for large m and n, it is thus Cauchy.
but,
$||x^m - x^n||_infty = lim_{p->infty} (int_0^1|x^m-x^n|^p dx)^{1/p} =_{| m<n} lim_{p->infty} (int_0^1(x^m-x^n)^p dx)^{1/p}$
I am not really sure how I could get this to show a contradiction, or if it's even the right way to go...
metric-spaces norm cauchy-sequences
asked Jan 24 at 18:05
bjornbjorn
1104
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In my text book, it is stated that for a sequence of functions ${x^n}$ in a metric space $C^0([0,1])$, it is Cauchy in $||-||_1$ norm but not in $||-||_infty$ norm because that would lead to a contradiction.
I am wondering what this contradiction is.
I tried expanding the norm:
$||x^m - x^n||_1 = int_0^1|x^m-x^n| dx =_{| m<n} int_0^1(x^m-x^n) dx = frac{1}{m+1} - frac{1}{n+1}$, which goes to zero (which is in C) for large m and n, it is thus Cauchy.
but,
$||x^m - x^n||_infty = lim_{p->infty} (int_0^1|x^m-x^n|^p dx)^{1/p} =_{| m<n} lim_{p->infty} (int_0^1(x^m-x^n)^p dx)^{1/p}$
I am not really sure how I could get this to show a contradiction, or if it's even the right way to go...
metric-spaces norm cauchy-sequences
asked Jan 24 at 18:05
bjornbjorn
1104
$endgroup$
1
$begingroup$
If it were cauchy (hence convergent) in $L^infty$, then it would be a sequence of continuous functions converging uniformly. So the limit has to be continuous, and at the same time equal to the point wise limit which is not continuous.
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– Calvin Khor
Jan 24 at 18:12
1
$begingroup$
You should work with $lVert f rVert_infty = sup_{t in [0,1]} lvert f(t) rvert$.
$endgroup$
– Paul Frost
Jan 24 at 21:16
add a comment |
$begingroup$
In my text book, it is stated that for a sequence of functions ${x^n}$ in a metric space $C^0([0,1])$, it is Cauchy in $||-||_1$ norm but not in $||-||_infty$ norm because that would lead to a contradiction.
I am wondering what this contradiction is.
I tried expanding the norm:
$||x^m - x^n||_1 = int_0^1|x^m-x^n| dx =_{| m<n} int_0^1(x^m-x^n) dx = frac{1}{m+1} - frac{1}{n+1}$, which goes to zero (which is in C) for large m and n, it is thus Cauchy.
but,
$||x^m - x^n||_infty = lim_{p->infty} (int_0^1|x^m-x^n|^p dx)^{1/p} =_{| m<n} lim_{p->infty} (int_0^1(x^m-x^n)^p dx)^{1/p}$
I am not really sure how I could get this to show a contradiction, or if it's even the right way to go...
metric-spaces norm cauchy-sequences
asked Jan 24 at 18:05
bjornbjorn
1104
$endgroup$
In my text book, it is stated that for a sequence of functions ${x^n}$ in a metric space $C^0([0,1])$, it is Cauchy in $||-||_1$ norm but not in $||-||_infty$ norm because that would lead to a contradiction.
I am wondering what this contradiction is.
I tried expanding the norm:
$||x^m - x^n||_1 = int_0^1|x^m-x^n| dx =_{| m<n} int_0^1(x^m-x^n) dx = frac{1}{m+1} - frac{1}{n+1}$, which goes to zero (which is in C) for large m and n, it is thus Cauchy.
but,
$||x^m - x^n||_infty = lim_{p->infty} (int_0^1|x^m-x^n|^p dx)^{1/p} =_{| m<n} lim_{p->infty} (int_0^1(x^m-x^n)^p dx)^{1/p}$
I am not really sure how I could get this to show a contradiction, or if it's even the right way to go...
metric-spaces norm cauchy-sequences
metric-spaces norm cauchy-sequences
asked Jan 24 at 18:05
bjornbjorn
1104
asked Jan 24 at 18:05
bjornbjorn
1104
asked Jan 24 at 18:05
bjornbjorn
1104
asked Jan 24 at 18:05
bjornbjorn
1104
asked Jan 24 at 18:05
bjornbjorn
1104
1104
1
$begingroup$
If it were cauchy (hence convergent) in $L^infty$, then it would be a sequence of continuous functions converging uniformly. So the limit has to be continuous, and at the same time equal to the point wise limit which is not continuous.
$endgroup$
– Calvin Khor
Jan 24 at 18:12
1
$begingroup$
You should work with $lVert f rVert_infty = sup_{t in [0,1]} lvert f(t) rvert$.
$endgroup$
– Paul Frost
Jan 24 at 21:16
add a comment |
1
$begingroup$
If it were cauchy (hence convergent) in $L^infty$, then it would be a sequence of continuous functions converging uniformly. So the limit has to be continuous, and at the same time equal to the point wise limit which is not continuous.
$endgroup$
– Calvin Khor
Jan 24 at 18:12
1
$begingroup$
You should work with $lVert f rVert_infty = sup_{t in [0,1]} lvert f(t) rvert$.
$endgroup$
– Paul Frost
Jan 24 at 21:16
1
1
$begingroup$
If it were cauchy (hence convergent) in $L^infty$, then it would be a sequence of continuous functions converging uniformly. So the limit has to be continuous, and at the same time equal to the point wise limit which is not continuous.
$endgroup$
– Calvin Khor
Jan 24 at 18:12
$begingroup$
If it were cauchy (hence convergent) in $L^infty$, then it would be a sequence of continuous functions converging uniformly. So the limit has to be continuous, and at the same time equal to the point wise limit which is not continuous.
$endgroup$
– Calvin Khor
Jan 24 at 18:12
1
1
$begingroup$
You should work with $lVert f rVert_infty = sup_{t in [0,1]} lvert f(t) rvert$.
$endgroup$
– Paul Frost
Jan 24 at 21:16
$begingroup$
You should work with $lVert f rVert_infty = sup_{t in [0,1]} lvert f(t) rvert$.
$endgroup$
– Paul Frost
Jan 24 at 21:16
add a comment |
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If it were cauchy (hence convergent) in $L^infty$, then it would be a sequence of continuous functions converging uniformly. So the limit has to be continuous, and at the same time equal to the point wise limit which is not continuous.
$endgroup$
– Calvin Khor
Jan 24 at 18:12
1
$begingroup$
You should work with $lVert f rVert_infty = sup_{t in [0,1]} lvert f(t) rvert$.
$endgroup$
– Paul Frost
Jan 24 at 21:16