Function in $L^2$ that doesn't vanishing












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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







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  • $begingroup$
    no.............
    $endgroup$
    – mathworker21
    Jan 1 at 11:23
















0












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    no.............
    $endgroup$
    – mathworker21
    Jan 1 at 11:23














0












0








0





$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




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asked Jan 1 at 11:19









GustaveGustave

720211




720211












  • $begingroup$
    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




$begingroup$
no.............
$endgroup$
– mathworker21
Jan 1 at 11:23










1 Answer
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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 $f(x)ne 0, forall x in [0,1]setminus N$.






share|cite|improve this answer









$endgroup$









  • 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













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1 Answer
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1 Answer
1






active

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active

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active

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3












$begingroup$

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 $f(x)ne 0, forall x in [0,1]setminus N$.






share|cite|improve this answer









$endgroup$









  • 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


















3












$begingroup$

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 $f(x)ne 0, forall x in [0,1]setminus N$.






share|cite|improve this answer









$endgroup$









  • 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
















3












3








3





$begingroup$

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 $f(x)ne 0, forall x in [0,1]setminus N$.






share|cite|improve this answer









$endgroup$



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 $f(x)ne 0, forall x in [0,1]setminus N$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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
















  • 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




















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