Mixing
Function in $L^2$ that doesn't vanishing
$begingroup$
Is this statement makes a sense: $fin L^2(0,1)$ such that $f(x)ne 0 ,forall xin (0,1)$ ?
measure-theory lp-spaces almost-everywhere
asked Jan 1 at 11:19
GustaveGustave
720211
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add a comment |
$begingroup$
Is this statement makes a sense: $fin L^2(0,1)$ such that $f(x)ne 0 ,forall xin (0,1)$ ?
measure-theory lp-spaces almost-everywhere
asked Jan 1 at 11:19
GustaveGustave
720211
$endgroup$
$begingroup$
no.............
$endgroup$
– mathworker21
Jan 1 at 11:23
add a comment |
$begingroup$
Is this statement makes a sense: $fin L^2(0,1)$ such that $f(x)ne 0 ,forall xin (0,1)$ ?
measure-theory lp-spaces almost-everywhere
asked Jan 1 at 11:19
GustaveGustave
720211
$endgroup$
Is this statement makes a sense: $fin L^2(0,1)$ such that $f(x)ne 0 ,forall xin (0,1)$ ?
measure-theory lp-spaces almost-everywhere
measure-theory lp-spaces almost-everywhere
asked Jan 1 at 11:19
GustaveGustave
720211
asked Jan 1 at 11:19
GustaveGustave
720211
asked Jan 1 at 11:19
GustaveGustave
720211
asked Jan 1 at 11:19
GustaveGustave
720211
asked Jan 1 at 11:19
GustaveGustave
720211
720211
$begingroup$
no.............
$endgroup$
– mathworker21
Jan 1 at 11:23
add a comment |
$begingroup$
no.............
$endgroup$
– mathworker21
Jan 1 at 11:23
$begingroup$
no.............
$endgroup$
– mathworker21
Jan 1 at 11:23
$begingroup$
no.............
$endgroup$
– mathworker21
Jan 1 at 11:23
add a comment |
1 Answer
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$begingroup$
Not really. If $f$ was such a function, then define $g in L^2(0,1)$ as
$$g(x) = begin{cases} 0, text{if } x = frac12 \
f(x), text{ otherwise}end{cases}$$
Then $f = g$ almost everywhere so they are equal as elements of $L^2(0,1)$.
However, what does makes sense is to require that $f ne 0$ almost everywhere, i.e. there exists $N subseteq [0,1]$ such that $lambda(N) = 0$ and that $f(x)ne 0, forall x in [0,1]setminus N$.
answered Jan 1 at 11:26
mechanodroidmechanodroid
27.1k62446
$endgroup$
1
$begingroup$
I think it should be said that there is both a notion of $L^2$, the set of square integrable functions, and $L^2=(L^2,langlecdot,cdotrangle)$ the inner product space of square integrable functions. So the OP does make sense in the set, and also does not make sense in the inner product space.
$endgroup$
– InequalitiesEverywhere
Jan 1 at 11:49
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
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oldest
votes
$begingroup$
Not really. If $f$ was such a function, then define $g in L^2(0,1)$ as
$$g(x) = begin{cases} 0, text{if } x = frac12 \
f(x), text{ otherwise}end{cases}$$
Then $f = g$ almost everywhere so they are equal as elements of $L^2(0,1)$.
However, what does makes sense is to require that $f ne 0$ almost everywhere, i.e. there exists $N subseteq [0,1]$ such that $lambda(N) = 0$ and that $f(x)ne 0, forall x in [0,1]setminus N$.
answered Jan 1 at 11:26
mechanodroidmechanodroid
27.1k62446
$endgroup$
1
$begingroup$
I think it should be said that there is both a notion of $L^2$, the set of square integrable functions, and $L^2=(L^2,langlecdot,cdotrangle)$ the inner product space of square integrable functions. So the OP does make sense in the set, and also does not make sense in the inner product space.
$endgroup$
– InequalitiesEverywhere
Jan 1 at 11:49
add a comment |
$begingroup$
Not really. If $f$ was such a function, then define $g in L^2(0,1)$ as
$$g(x) = begin{cases} 0, text{if } x = frac12 \
f(x), text{ otherwise}end{cases}$$
Then $f = g$ almost everywhere so they are equal as elements of $L^2(0,1)$.
However, what does makes sense is to require that $f ne 0$ almost everywhere, i.e. there exists $N subseteq [0,1]$ such that $lambda(N) = 0$ and that $f(x)ne 0, forall x in [0,1]setminus N$.
answered Jan 1 at 11:26
mechanodroidmechanodroid
27.1k62446
$endgroup$
1
$begingroup$
I think it should be said that there is both a notion of $L^2$, the set of square integrable functions, and $L^2=(L^2,langlecdot,cdotrangle)$ the inner product space of square integrable functions. So the OP does make sense in the set, and also does not make sense in the inner product space.
$endgroup$
– InequalitiesEverywhere
Jan 1 at 11:49
add a comment |
$begingroup$
Not really. If $f$ was such a function, then define $g in L^2(0,1)$ as
$$g(x) = begin{cases} 0, text{if } x = frac12 \
f(x), text{ otherwise}end{cases}$$
Then $f = g$ almost everywhere so they are equal as elements of $L^2(0,1)$.
However, what does makes sense is to require that $f ne 0$ almost everywhere, i.e. there exists $N subseteq [0,1]$ such that $lambda(N) = 0$ and that $f(x)ne 0, forall x in [0,1]setminus N$.
answered Jan 1 at 11:26
mechanodroidmechanodroid
27.1k62446
$endgroup$
Not really. If $f$ was such a function, then define $g in L^2(0,1)$ as
$$g(x) = begin{cases} 0, text{if } x = frac12 \
f(x), text{ otherwise}end{cases}$$
Then $f = g$ almost everywhere so they are equal as elements of $L^2(0,1)$.
However, what does makes sense is to require that $f ne 0$ almost everywhere, i.e. there exists $N subseteq [0,1]$ such that $lambda(N) = 0$ and that $f(x)ne 0, forall x in [0,1]setminus N$.
answered Jan 1 at 11:26
mechanodroidmechanodroid
27.1k62446
answered Jan 1 at 11:26
mechanodroidmechanodroid
27.1k62446
answered Jan 1 at 11:26
mechanodroidmechanodroid
27.1k62446
answered Jan 1 at 11:26
mechanodroidmechanodroid
27.1k62446
27.1k62446
1
$begingroup$
I think it should be said that there is both a notion of $L^2$, the set of square integrable functions, and $L^2=(L^2,langlecdot,cdotrangle)$ the inner product space of square integrable functions. So the OP does make sense in the set, and also does not make sense in the inner product space.
$endgroup$
– InequalitiesEverywhere
Jan 1 at 11:49
add a comment |
1
$begingroup$
I think it should be said that there is both a notion of $L^2$, the set of square integrable functions, and $L^2=(L^2,langlecdot,cdotrangle)$ the inner product space of square integrable functions. So the OP does make sense in the set, and also does not make sense in the inner product space.
$endgroup$
– InequalitiesEverywhere
Jan 1 at 11:49
1
1
$begingroup$
I think it should be said that there is both a notion of $L^2$, the set of square integrable functions, and $L^2=(L^2,langlecdot,cdotrangle)$ the inner product space of square integrable functions. So the OP does make sense in the set, and also does not make sense in the inner product space.
$endgroup$
– InequalitiesEverywhere
Jan 1 at 11:49
$begingroup$
I think it should be said that there is both a notion of $L^2$, the set of square integrable functions, and $L^2=(L^2,langlecdot,cdotrangle)$ the inner product space of square integrable functions. So the OP does make sense in the set, and also does not make sense in the inner product space.
$endgroup$
– InequalitiesEverywhere
Jan 1 at 11:49
add a comment |
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$begingroup$
no.............
$endgroup$
– mathworker21
Jan 1 at 11:23