Is the Set of Continuous Functions that are the Sum of Even and Odd Functions Meager?
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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
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up vote
1
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
add a comment |
up vote
1
down vote
favorite
up vote
1
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
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 37 mins ago
Gaby Boy Analysis
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1 Answer
1
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up vote
4
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.)
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
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.)
add a comment |
up vote
4
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.)
add a comment |
up vote
4
down vote
accepted
up vote
4
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.)
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.)
edited 28 mins ago
answered 34 mins ago
copper.hat
125k558158
125k558158
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