Is the Set of Continuous Functions that are the Sum of Even and Odd Functions Meager?
up vote
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Consider $X = mathcal{C}([−1,1])$ with the usual norm $|f|_{infty} = sup_{tin [−1,1]}|f(t)|.$
Define
$$mathcal{A}_{+}={ f in X : f(t)=f(−t) space forall tin [−1,1]},$$
$$mathcal{A}_{−}={ f in X : f(t)=−f(−t) space forall t in [−1,1]}. $$
Is $mathcal{A}_{+} +mathcal{A}_{−} = {f +g : f in mathcal{A}_{+},g in mathcal{A}_{−}}$ meager?
I know this set is dense by the Stone-Weierstrass Theorem. However, that doesn't really help. I also know that if the set is closed, then it is meager, but I have difficulties deciding whether it is closed or not. I know the exponential function is a limit of a sequence of a sum of even and odd functions, however one could define it to be that, in which case it doesn't help.
Any hints on how to get going on this problem, and on whether the set $mathcal{A}_{+}+{A}_{-} $ is closed or not? Thank you in advance.
real-analysis general-topology functional-analysis metric-spaces baire-category
asked Dec 2 at 7:14
Gaby Alfonso
670315
add a comment |
up vote
6
down vote
favorite
Consider $X = mathcal{C}([−1,1])$ with the usual norm $|f|_{infty} = sup_{tin [−1,1]}|f(t)|.$
Define
$$mathcal{A}_{+}={ f in X : f(t)=f(−t) space forall tin [−1,1]},$$
$$mathcal{A}_{−}={ f in X : f(t)=−f(−t) space forall t in [−1,1]}. $$
Is $mathcal{A}_{+} +mathcal{A}_{−} = {f +g : f in mathcal{A}_{+},g in mathcal{A}_{−}}$ meager?
I know this set is dense by the Stone-Weierstrass Theorem. However, that doesn't really help. I also know that if the set is closed, then it is meager, but I have difficulties deciding whether it is closed or not. I know the exponential function is a limit of a sequence of a sum of even and odd functions, however one could define it to be that, in which case it doesn't help.
Any hints on how to get going on this problem, and on whether the set $mathcal{A}_{+}+{A}_{-} $ is closed or not? Thank you in advance.
real-analysis general-topology functional-analysis metric-spaces baire-category
asked Dec 2 at 7:14
Gaby Alfonso
670315
add a comment |
up vote
6
down vote
favorite
up vote
6
down vote
favorite
Consider $X = mathcal{C}([−1,1])$ with the usual norm $|f|_{infty} = sup_{tin [−1,1]}|f(t)|.$
Define
$$mathcal{A}_{+}={ f in X : f(t)=f(−t) space forall tin [−1,1]},$$
$$mathcal{A}_{−}={ f in X : f(t)=−f(−t) space forall t in [−1,1]}. $$
Is $mathcal{A}_{+} +mathcal{A}_{−} = {f +g : f in mathcal{A}_{+},g in mathcal{A}_{−}}$ meager?
I know this set is dense by the Stone-Weierstrass Theorem. However, that doesn't really help. I also know that if the set is closed, then it is meager, but I have difficulties deciding whether it is closed or not. I know the exponential function is a limit of a sequence of a sum of even and odd functions, however one could define it to be that, in which case it doesn't help.
Any hints on how to get going on this problem, and on whether the set $mathcal{A}_{+}+{A}_{-} $ is closed or not? Thank you in advance.
real-analysis general-topology functional-analysis metric-spaces baire-category
asked Dec 2 at 7:14
Gaby Alfonso
670315
Consider $X = mathcal{C}([−1,1])$ with the usual norm $|f|_{infty} = sup_{tin [−1,1]}|f(t)|.$
Define
$$mathcal{A}_{+}={ f in X : f(t)=f(−t) space forall tin [−1,1]},$$
$$mathcal{A}_{−}={ f in X : f(t)=−f(−t) space forall t in [−1,1]}. $$
Is $mathcal{A}_{+} +mathcal{A}_{−} = {f +g : f in mathcal{A}_{+},g in mathcal{A}_{−}}$ meager?
I know this set is dense by the Stone-Weierstrass Theorem. However, that doesn't really help. I also know that if the set is closed, then it is meager, but I have difficulties deciding whether it is closed or not. I know the exponential function is a limit of a sequence of a sum of even and odd functions, however one could define it to be that, in which case it doesn't help.
Any hints on how to get going on this problem, and on whether the set $mathcal{A}_{+}+{A}_{-} $ is closed or not? Thank you in advance.
real-analysis general-topology functional-analysis metric-spaces baire-category
real-analysis general-topology functional-analysis metric-spaces baire-category
asked Dec 2 at 7:14
Gaby Alfonso
670315
asked Dec 2 at 7:14
Gaby Alfonso
670315
asked Dec 2 at 7:14
Gaby Alfonso
670315
asked Dec 2 at 7:14
Gaby Alfonso
670315
asked Dec 2 at 7:14
Gaby Alfonso
670315
670315
add a comment |
add a comment |
1 Answer
1
active
oldest
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up vote
16
down vote
accepted
Note that any function can be written as
$f(x) = {1 over 2} (f(x) + f(-x)) + {1 over 2} (f(x) - f(-x)) $, so
$mathcal{A}_{+} +mathcal{A}_{−} = X$, which is not meagre.
(It is not meagre because $C[-1,1]$ is a complete metric space.)
answered Dec 2 at 7:17
copper.hat
125k559159
1
As an aside, if OP is familiar with Fourier series this decomposition should be part of the associated intuition.
– R..
Dec 2 at 14:07
@R.. Fourier decomposition is way more complicated than necessary.
– leftaroundabout
Dec 2 at 19:05
@leftaroundabout: Absolutely. copper.hat's answer is simpler and more direct, but having studied Fourier series should leave one with an intuition that there is a natural decomposition into even and odd parts, whereas otherwise this may seem like a surprising and/or obscure result.
– R..
Dec 2 at 21:18
@R.. hm, I rather see it the other way around – this decomposition is one of the foundations that make Fourier analysis possible. But, I suppose this is somewhat a matter of taste.
– leftaroundabout
Dec 2 at 21:50
1
I guess I would see the decomposition of square matrices into symmetric and skew symmetric parts a more natural intuition. Fourier series requires a lot more baggage.
– copper.hat
Dec 2 at 22:36
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
16
down vote
accepted
Note that any function can be written as
$f(x) = {1 over 2} (f(x) + f(-x)) + {1 over 2} (f(x) - f(-x)) $, so
$mathcal{A}_{+} +mathcal{A}_{−} = X$, which is not meagre.
(It is not meagre because $C[-1,1]$ is a complete metric space.)
answered Dec 2 at 7:17
copper.hat
125k559159
1
As an aside, if OP is familiar with Fourier series this decomposition should be part of the associated intuition.
– R..
Dec 2 at 14:07
@R.. Fourier decomposition is way more complicated than necessary.
– leftaroundabout
Dec 2 at 19:05
@leftaroundabout: Absolutely. copper.hat's answer is simpler and more direct, but having studied Fourier series should leave one with an intuition that there is a natural decomposition into even and odd parts, whereas otherwise this may seem like a surprising and/or obscure result.
– R..
Dec 2 at 21:18
@R.. hm, I rather see it the other way around – this decomposition is one of the foundations that make Fourier analysis possible. But, I suppose this is somewhat a matter of taste.
– leftaroundabout
Dec 2 at 21:50
1
I guess I would see the decomposition of square matrices into symmetric and skew symmetric parts a more natural intuition. Fourier series requires a lot more baggage.
– copper.hat
Dec 2 at 22:36
add a comment |
up vote
16
down vote
accepted
Note that any function can be written as
$f(x) = {1 over 2} (f(x) + f(-x)) + {1 over 2} (f(x) - f(-x)) $, so
$mathcal{A}_{+} +mathcal{A}_{−} = X$, which is not meagre.
(It is not meagre because $C[-1,1]$ is a complete metric space.)
answered Dec 2 at 7:17
copper.hat
125k559159
1
As an aside, if OP is familiar with Fourier series this decomposition should be part of the associated intuition.
– R..
Dec 2 at 14:07
@R.. Fourier decomposition is way more complicated than necessary.
– leftaroundabout
Dec 2 at 19:05
@leftaroundabout: Absolutely. copper.hat's answer is simpler and more direct, but having studied Fourier series should leave one with an intuition that there is a natural decomposition into even and odd parts, whereas otherwise this may seem like a surprising and/or obscure result.
– R..
Dec 2 at 21:18
@R.. hm, I rather see it the other way around – this decomposition is one of the foundations that make Fourier analysis possible. But, I suppose this is somewhat a matter of taste.
– leftaroundabout
Dec 2 at 21:50
1
I guess I would see the decomposition of square matrices into symmetric and skew symmetric parts a more natural intuition. Fourier series requires a lot more baggage.
– copper.hat
Dec 2 at 22:36
add a comment |
up vote
16
down vote
accepted
up vote
16
down vote
accepted
Note that any function can be written as
$f(x) = {1 over 2} (f(x) + f(-x)) + {1 over 2} (f(x) - f(-x)) $, so
$mathcal{A}_{+} +mathcal{A}_{−} = X$, which is not meagre.
(It is not meagre because $C[-1,1]$ is a complete metric space.)
answered Dec 2 at 7:17
copper.hat
125k559159
Note that any function can be written as
$f(x) = {1 over 2} (f(x) + f(-x)) + {1 over 2} (f(x) - f(-x)) $, so
$mathcal{A}_{+} +mathcal{A}_{−} = X$, which is not meagre.
(It is not meagre because $C[-1,1]$ is a complete metric space.)
answered Dec 2 at 7:17
copper.hat
125k559159
edited Dec 2 at 7:23
answered Dec 2 at 7:17
copper.hat
125k559159
answered Dec 2 at 7:17
copper.hat
125k559159
answered Dec 2 at 7:17
copper.hat
125k559159
125k559159
1
As an aside, if OP is familiar with Fourier series this decomposition should be part of the associated intuition.
– R..
Dec 2 at 14:07
@R.. Fourier decomposition is way more complicated than necessary.
– leftaroundabout
Dec 2 at 19:05
@leftaroundabout: Absolutely. copper.hat's answer is simpler and more direct, but having studied Fourier series should leave one with an intuition that there is a natural decomposition into even and odd parts, whereas otherwise this may seem like a surprising and/or obscure result.
– R..
Dec 2 at 21:18
@R.. hm, I rather see it the other way around – this decomposition is one of the foundations that make Fourier analysis possible. But, I suppose this is somewhat a matter of taste.
– leftaroundabout
Dec 2 at 21:50
1
I guess I would see the decomposition of square matrices into symmetric and skew symmetric parts a more natural intuition. Fourier series requires a lot more baggage.
– copper.hat
Dec 2 at 22:36
add a comment |
1
As an aside, if OP is familiar with Fourier series this decomposition should be part of the associated intuition.
– R..
Dec 2 at 14:07
@R.. Fourier decomposition is way more complicated than necessary.
– leftaroundabout
Dec 2 at 19:05
@leftaroundabout: Absolutely. copper.hat's answer is simpler and more direct, but having studied Fourier series should leave one with an intuition that there is a natural decomposition into even and odd parts, whereas otherwise this may seem like a surprising and/or obscure result.
– R..
Dec 2 at 21:18
@R.. hm, I rather see it the other way around – this decomposition is one of the foundations that make Fourier analysis possible. But, I suppose this is somewhat a matter of taste.
– leftaroundabout
Dec 2 at 21:50
1
I guess I would see the decomposition of square matrices into symmetric and skew symmetric parts a more natural intuition. Fourier series requires a lot more baggage.
– copper.hat
Dec 2 at 22:36
1
1
As an aside, if OP is familiar with Fourier series this decomposition should be part of the associated intuition.
– R..
Dec 2 at 14:07
As an aside, if OP is familiar with Fourier series this decomposition should be part of the associated intuition.
– R..
Dec 2 at 14:07
@R.. Fourier decomposition is way more complicated than necessary.
– leftaroundabout
Dec 2 at 19:05
@R.. Fourier decomposition is way more complicated than necessary.
– leftaroundabout
Dec 2 at 19:05
@leftaroundabout: Absolutely. copper.hat's answer is simpler and more direct, but having studied Fourier series should leave one with an intuition that there is a natural decomposition into even and odd parts, whereas otherwise this may seem like a surprising and/or obscure result.
– R..
Dec 2 at 21:18
@leftaroundabout: Absolutely. copper.hat's answer is simpler and more direct, but having studied Fourier series should leave one with an intuition that there is a natural decomposition into even and odd parts, whereas otherwise this may seem like a surprising and/or obscure result.
– R..
Dec 2 at 21:18
@R.. hm, I rather see it the other way around – this decomposition is one of the foundations that make Fourier analysis possible. But, I suppose this is somewhat a matter of taste.
– leftaroundabout
Dec 2 at 21:50
@R.. hm, I rather see it the other way around – this decomposition is one of the foundations that make Fourier analysis possible. But, I suppose this is somewhat a matter of taste.
– leftaroundabout
Dec 2 at 21:50
1
1
I guess I would see the decomposition of square matrices into symmetric and skew symmetric parts a more natural intuition. Fourier series requires a lot more baggage.
– copper.hat
Dec 2 at 22:36
I guess I would see the decomposition of square matrices into symmetric and skew symmetric parts a more natural intuition. Fourier series requires a lot more baggage.
– copper.hat
Dec 2 at 22:36
add a comment |
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