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Show that there exists no $f_0in X$ such that $||f_0||=1$ and $textdist(f_0,Y)=1$
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Let $X=fin C[0,1]:f(0)=0$ with the sup-norm and $Y=fin X:int _0^1 f=0$
Show that there exists no $f_0in X$ such that $||f_0||=1$ and $textdist(f_0,Y)=1$.
Suppose such a $f_0$ exists then $||f_0||=1implies sup f_0(x):xin [0,1]=1$ and $inff_0-f=1$
Then $forall f $ with $int f=0$ we have $sup ge 1$
Then $forall f $ with $int f=0$ there exists $x_0$ depending on $f$ such that $|f_0(x)-f(x)|ge 1$
Also $sup =1implies exists x_0$ such that $f_0(x_0)=1$.
I am unable to use the above information to find such a $f_0$ does not exist.Can someone please help?
calculus functional-analysis functions continuity
asked Aug 6 at 2:40
PureMathematics
325
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Let $X=fin C[0,1]:f(0)=0$ with the sup-norm and $Y=fin X:int _0^1 f=0$
Show that there exists no $f_0in X$ such that $||f_0||=1$ and $textdist(f_0,Y)=1$.
Suppose such a $f_0$ exists then $||f_0||=1implies sup f_0(x):xin [0,1]=1$ and $inff_0-f=1$
Then $forall f $ with $int f=0$ we have $sup ge 1$
Then $forall f $ with $int f=0$ there exists $x_0$ depending on $f$ such that $|f_0(x)-f(x)|ge 1$
Also $sup =1implies exists x_0$ such that $f_0(x_0)=1$.
I am unable to use the above information to find such a $f_0$ does not exist.Can someone please help?
calculus functional-analysis functions continuity
asked Aug 6 at 2:40
PureMathematics
325
Note that since $0 in Y$, we have $textdist(f_0,Y) le |f_0-0| = 1$.
– copper.hat
Aug 6 at 4:24
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up vote
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Let $X=fin C[0,1]:f(0)=0$ with the sup-norm and $Y=fin X:int _0^1 f=0$
Show that there exists no $f_0in X$ such that $||f_0||=1$ and $textdist(f_0,Y)=1$.
Suppose such a $f_0$ exists then $||f_0||=1implies sup f_0(x):xin [0,1]=1$ and $inff_0-f=1$
Then $forall f $ with $int f=0$ we have $sup ge 1$
Then $forall f $ with $int f=0$ there exists $x_0$ depending on $f$ such that $|f_0(x)-f(x)|ge 1$
Also $sup =1implies exists x_0$ such that $f_0(x_0)=1$.
I am unable to use the above information to find such a $f_0$ does not exist.Can someone please help?
calculus functional-analysis functions continuity
asked Aug 6 at 2:40
PureMathematics
325
Let $X=fin C[0,1]:f(0)=0$ with the sup-norm and $Y=fin X:int _0^1 f=0$
Show that there exists no $f_0in X$ such that $||f_0||=1$ and $textdist(f_0,Y)=1$.
Suppose such a $f_0$ exists then $||f_0||=1implies sup f_0(x):xin [0,1]=1$ and $inff_0-f=1$
Then $forall f $ with $int f=0$ we have $sup ge 1$
Then $forall f $ with $int f=0$ there exists $x_0$ depending on $f$ such that $|f_0(x)-f(x)|ge 1$
Also $sup =1implies exists x_0$ such that $f_0(x_0)=1$.
I am unable to use the above information to find such a $f_0$ does not exist.Can someone please help?
calculus functional-analysis functions continuity
asked Aug 6 at 2:40
PureMathematics
325
asked Aug 6 at 2:40
PureMathematics
325
asked Aug 6 at 2:40
PureMathematics
325
asked Aug 6 at 2:40
PureMathematics
325
325
Note that since $0 in Y$, we have $textdist(f_0,Y) le |f_0-0| = 1$.
– copper.hat
Aug 6 at 4:24
add a comment |Â
Note that since $0 in Y$, we have $textdist(f_0,Y) le |f_0-0| = 1$.
– copper.hat
Aug 6 at 4:24
Note that since $0 in Y$, we have $textdist(f_0,Y) le |f_0-0| = 1$.
– copper.hat
Aug 6 at 4:24
Note that since $0 in Y$, we have $textdist(f_0,Y) le |f_0-0| = 1$.
– copper.hat
Aug 6 at 4:24
add a comment |Â
2 Answers
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Luiz Cordeiro's answer is nice and very natural, but you can get a broader perspective and avoid too much actual fiddling around with functions by thinking more abstractly. Let $I:XtomathbbR$ be the functional $I(f)=int_0^1 f$. Note that $Y=ker(f)$ so $X/Y$ is a one-dimensional Banach space (since $I$ induces a linear isomorphism $X/YcongmathbbR$), with the usual quotient norm $|f+Y|=operatornamedist(f,Y)$. But since $X/Y$ is one-dimensional, there is only one possible norm on it up to scaling, and in particular the functional $I$ must preserve the norm up to scaling. That is, there is some constant $c$ such that $$|I(f)|=coperatornamedist(f,Y)$$ for all $fin X$.
Now, if we can just prove that $cgeq 1$, we're done. Indeed, that implies that $operatornamedist(f,Y)leq |I(f)|$ for any $fin X$. If $|f|=1$, then $|I(f)|<1$, with the inequality being strict because $f(0)$ must be $0$. So, for any $fin X$ with $|f|=1$ we conclude that $operatornamedist(f,Y)<1$.
So we just need to prove that $cgeq 1$. But since the same constant $c$ works for all $fin X$, we can prove this by just considering a single nice function $f$. For instance, you could take $f(t)=t$. Then $I(f)=1/2$, so it suffices to find, for any $epsilon>0$, a function $g_epsilonin Y$ such that $|f-g_epsilon|leq 1/2+epsilon$. This is not so hard to just construct concretely, by fiddling around with piecewise linear functions. For instance, you could observe that $g(t)=t-1/2$ would have the desired properties except that $g(0)neq 0$, and then modify this function slightly to get the desired $g_epsilon$ (let $g_epsilon(0)=0$ and then have $g_epsilon(t)$ very quickly jump down to close to $-1/2$, and thereafter let $g_epsilon(t)$ be slightly smaller than $g(t)$ to make the integral of $g_epsilon$ balance out to $0$).
(In fact, $c=1$, so $operatornamedist(f,Y)$ is just $|I(f)|$. I'll leave it as an exercise to prove the other direction $cleq 1$, which is easier.)
answered Aug 6 at 5:27
Eric Wofsey
163k12190301
+1: Nice approach using quotient space.
– copper.hat
Aug 6 at 15:09
add a comment |Â
up vote
1
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You can prove that if $Vert f_0Vert=1$, then there is $fin Y$ with $Vert f_0-fVert<1$. Below is a way to do it.
Since $f_0(0)=0$, then there is an interval $[0,epsilon)$ around zero on which $|f_0|<1/2$. Then $|f_0(epsilon)|leq1/2$.
In order to guarantee that the function $f$ we will construct will satisfy $Vert f-f_0Vert<1$, we just make $f$ "a little positive" when $f_0=1$ and "a little negative" when $f_0=-1$. So choose a small, but positive, $delta$ (how small will be specified later). Let $n:[-1,1]to[-delta,delta]$ be given by
$$n(x)=begincases
x+1-delta&text if -1leq xleq -1+delta\
0&text if -1+deltaleq xleq 1-delta\
x-1+delta&text if 1-deltaleq xleq 1endcases$$
This function $n$ is "a little positive around $1$" and "a little negative" around $-1$. Anyway, define $f(x)=n(f_0(x))$ for all $xin[epsilon,1]$. You can then prove that $|f(x)-f_0(x)|leq 1-delta$ for all $xin[epsilon,1]$.
As long as $delta<1/2$, we also have $f(epsilon)=0$. I'll assume this from now on, because $delta$ can be initially taken very small.
Now we need to define $f$ on $[0,epsilon]$, in a way that $fin Y$. We will define $f$ as a "spike". If we choose $cinmathbbR$, define $f:[0,epsilon]tomathbbR$ in such a way that:
- $f$ is linear on $[0,epsilon/2]$;
- $f$ is linear on $[epsilon/2,epsilon]$
- $f(epsilon/2)=c$.
Then the integral $int_0^epsilon f(x)dx$ is simply the area of the triangle of basis $epsilon$ and heigth $c$, so it is equal to $cepsilon/2$.
We need to guarantee that $int_0^1f(x)dx=0$. So we want that
beginalign*
int_0^epsilon f(x)dx+int_epsilon^1f(x)dx&=0\
epsilon c/2=-int_epsilon^1f(x)dx
endalign*
thus we choose
$$c=-2int_epsilon^1f(x)dx/epsilon$$
Note that
$$|c|leq 2int_epsilon^1|f(x)|dxleq 2delta/epsilon$$
because of the definition of $f=ncirc f_0$ on $[epsilon,1]$, and $|n|leqdelta$. Just choose $delta$ small enough (depending only on $epsilon$) in such a way that $|c|<1/4$.
Repeating all we did so far: as long as we take $delta$ small enough (namely, $delta<1/2$ and $delta<epsilon/8$), we have defined $f$ on all of $[0,1]$, in such a way that
- $fin Y$ (this is precisely why we take $c$ as above).
- for all $xin[epsilon,1]$, $|f(x)-f_0(x)|leq 1-delta$ (this is a computation we did above);
- for all $xin[0,epsilon]$,
$$|f(x)-f_0(x)|leq |f(x)|+|f_0(x)|leq 1/4+1/2=3/4$$
(this is the choice of $epsilon$ and $c$.)
The second and third items above imply that $Vert f-f_0Vertleq max(1-delta,3/4)$, which is smaller than $1$.
answered Aug 6 at 4:33
Luiz Cordeiro
12.2k1041
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2 Answers
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active
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Luiz Cordeiro's answer is nice and very natural, but you can get a broader perspective and avoid too much actual fiddling around with functions by thinking more abstractly. Let $I:XtomathbbR$ be the functional $I(f)=int_0^1 f$. Note that $Y=ker(f)$ so $X/Y$ is a one-dimensional Banach space (since $I$ induces a linear isomorphism $X/YcongmathbbR$), with the usual quotient norm $|f+Y|=operatornamedist(f,Y)$. But since $X/Y$ is one-dimensional, there is only one possible norm on it up to scaling, and in particular the functional $I$ must preserve the norm up to scaling. That is, there is some constant $c$ such that $$|I(f)|=coperatornamedist(f,Y)$$ for all $fin X$.
Now, if we can just prove that $cgeq 1$, we're done. Indeed, that implies that $operatornamedist(f,Y)leq |I(f)|$ for any $fin X$. If $|f|=1$, then $|I(f)|<1$, with the inequality being strict because $f(0)$ must be $0$. So, for any $fin X$ with $|f|=1$ we conclude that $operatornamedist(f,Y)<1$.
So we just need to prove that $cgeq 1$. But since the same constant $c$ works for all $fin X$, we can prove this by just considering a single nice function $f$. For instance, you could take $f(t)=t$. Then $I(f)=1/2$, so it suffices to find, for any $epsilon>0$, a function $g_epsilonin Y$ such that $|f-g_epsilon|leq 1/2+epsilon$. This is not so hard to just construct concretely, by fiddling around with piecewise linear functions. For instance, you could observe that $g(t)=t-1/2$ would have the desired properties except that $g(0)neq 0$, and then modify this function slightly to get the desired $g_epsilon$ (let $g_epsilon(0)=0$ and then have $g_epsilon(t)$ very quickly jump down to close to $-1/2$, and thereafter let $g_epsilon(t)$ be slightly smaller than $g(t)$ to make the integral of $g_epsilon$ balance out to $0$).
(In fact, $c=1$, so $operatornamedist(f,Y)$ is just $|I(f)|$. I'll leave it as an exercise to prove the other direction $cleq 1$, which is easier.)
answered Aug 6 at 5:27
Eric Wofsey
163k12190301
+1: Nice approach using quotient space.
– copper.hat
Aug 6 at 15:09
add a comment |Â
up vote
2
down vote
Luiz Cordeiro's answer is nice and very natural, but you can get a broader perspective and avoid too much actual fiddling around with functions by thinking more abstractly. Let $I:XtomathbbR$ be the functional $I(f)=int_0^1 f$. Note that $Y=ker(f)$ so $X/Y$ is a one-dimensional Banach space (since $I$ induces a linear isomorphism $X/YcongmathbbR$), with the usual quotient norm $|f+Y|=operatornamedist(f,Y)$. But since $X/Y$ is one-dimensional, there is only one possible norm on it up to scaling, and in particular the functional $I$ must preserve the norm up to scaling. That is, there is some constant $c$ such that $$|I(f)|=coperatornamedist(f,Y)$$ for all $fin X$.
Now, if we can just prove that $cgeq 1$, we're done. Indeed, that implies that $operatornamedist(f,Y)leq |I(f)|$ for any $fin X$. If $|f|=1$, then $|I(f)|<1$, with the inequality being strict because $f(0)$ must be $0$. So, for any $fin X$ with $|f|=1$ we conclude that $operatornamedist(f,Y)<1$.
So we just need to prove that $cgeq 1$. But since the same constant $c$ works for all $fin X$, we can prove this by just considering a single nice function $f$. For instance, you could take $f(t)=t$. Then $I(f)=1/2$, so it suffices to find, for any $epsilon>0$, a function $g_epsilonin Y$ such that $|f-g_epsilon|leq 1/2+epsilon$. This is not so hard to just construct concretely, by fiddling around with piecewise linear functions. For instance, you could observe that $g(t)=t-1/2$ would have the desired properties except that $g(0)neq 0$, and then modify this function slightly to get the desired $g_epsilon$ (let $g_epsilon(0)=0$ and then have $g_epsilon(t)$ very quickly jump down to close to $-1/2$, and thereafter let $g_epsilon(t)$ be slightly smaller than $g(t)$ to make the integral of $g_epsilon$ balance out to $0$).
(In fact, $c=1$, so $operatornamedist(f,Y)$ is just $|I(f)|$. I'll leave it as an exercise to prove the other direction $cleq 1$, which is easier.)
answered Aug 6 at 5:27
Eric Wofsey
163k12190301
+1: Nice approach using quotient space.
– copper.hat
Aug 6 at 15:09
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Luiz Cordeiro's answer is nice and very natural, but you can get a broader perspective and avoid too much actual fiddling around with functions by thinking more abstractly. Let $I:XtomathbbR$ be the functional $I(f)=int_0^1 f$. Note that $Y=ker(f)$ so $X/Y$ is a one-dimensional Banach space (since $I$ induces a linear isomorphism $X/YcongmathbbR$), with the usual quotient norm $|f+Y|=operatornamedist(f,Y)$. But since $X/Y$ is one-dimensional, there is only one possible norm on it up to scaling, and in particular the functional $I$ must preserve the norm up to scaling. That is, there is some constant $c$ such that $$|I(f)|=coperatornamedist(f,Y)$$ for all $fin X$.
Now, if we can just prove that $cgeq 1$, we're done. Indeed, that implies that $operatornamedist(f,Y)leq |I(f)|$ for any $fin X$. If $|f|=1$, then $|I(f)|<1$, with the inequality being strict because $f(0)$ must be $0$. So, for any $fin X$ with $|f|=1$ we conclude that $operatornamedist(f,Y)<1$.
So we just need to prove that $cgeq 1$. But since the same constant $c$ works for all $fin X$, we can prove this by just considering a single nice function $f$. For instance, you could take $f(t)=t$. Then $I(f)=1/2$, so it suffices to find, for any $epsilon>0$, a function $g_epsilonin Y$ such that $|f-g_epsilon|leq 1/2+epsilon$. This is not so hard to just construct concretely, by fiddling around with piecewise linear functions. For instance, you could observe that $g(t)=t-1/2$ would have the desired properties except that $g(0)neq 0$, and then modify this function slightly to get the desired $g_epsilon$ (let $g_epsilon(0)=0$ and then have $g_epsilon(t)$ very quickly jump down to close to $-1/2$, and thereafter let $g_epsilon(t)$ be slightly smaller than $g(t)$ to make the integral of $g_epsilon$ balance out to $0$).
(In fact, $c=1$, so $operatornamedist(f,Y)$ is just $|I(f)|$. I'll leave it as an exercise to prove the other direction $cleq 1$, which is easier.)
answered Aug 6 at 5:27
Eric Wofsey
163k12190301
Luiz Cordeiro's answer is nice and very natural, but you can get a broader perspective and avoid too much actual fiddling around with functions by thinking more abstractly. Let $I:XtomathbbR$ be the functional $I(f)=int_0^1 f$. Note that $Y=ker(f)$ so $X/Y$ is a one-dimensional Banach space (since $I$ induces a linear isomorphism $X/YcongmathbbR$), with the usual quotient norm $|f+Y|=operatornamedist(f,Y)$. But since $X/Y$ is one-dimensional, there is only one possible norm on it up to scaling, and in particular the functional $I$ must preserve the norm up to scaling. That is, there is some constant $c$ such that $$|I(f)|=coperatornamedist(f,Y)$$ for all $fin X$.
Now, if we can just prove that $cgeq 1$, we're done. Indeed, that implies that $operatornamedist(f,Y)leq |I(f)|$ for any $fin X$. If $|f|=1$, then $|I(f)|<1$, with the inequality being strict because $f(0)$ must be $0$. So, for any $fin X$ with $|f|=1$ we conclude that $operatornamedist(f,Y)<1$.
So we just need to prove that $cgeq 1$. But since the same constant $c$ works for all $fin X$, we can prove this by just considering a single nice function $f$. For instance, you could take $f(t)=t$. Then $I(f)=1/2$, so it suffices to find, for any $epsilon>0$, a function $g_epsilonin Y$ such that $|f-g_epsilon|leq 1/2+epsilon$. This is not so hard to just construct concretely, by fiddling around with piecewise linear functions. For instance, you could observe that $g(t)=t-1/2$ would have the desired properties except that $g(0)neq 0$, and then modify this function slightly to get the desired $g_epsilon$ (let $g_epsilon(0)=0$ and then have $g_epsilon(t)$ very quickly jump down to close to $-1/2$, and thereafter let $g_epsilon(t)$ be slightly smaller than $g(t)$ to make the integral of $g_epsilon$ balance out to $0$).
(In fact, $c=1$, so $operatornamedist(f,Y)$ is just $|I(f)|$. I'll leave it as an exercise to prove the other direction $cleq 1$, which is easier.)
answered Aug 6 at 5:27
Eric Wofsey
163k12190301
edited Aug 6 at 5:33
answered Aug 6 at 5:27
Eric Wofsey
163k12190301
answered Aug 6 at 5:27
Eric Wofsey
163k12190301
answered Aug 6 at 5:27
Eric Wofsey
163k12190301
163k12190301
+1: Nice approach using quotient space.
– copper.hat
Aug 6 at 15:09
add a comment |Â
+1: Nice approach using quotient space.
– copper.hat
Aug 6 at 15:09
+1: Nice approach using quotient space.
– copper.hat
Aug 6 at 15:09
+1: Nice approach using quotient space.
– copper.hat
Aug 6 at 15:09
add a comment |Â
up vote
1
down vote
You can prove that if $Vert f_0Vert=1$, then there is $fin Y$ with $Vert f_0-fVert<1$. Below is a way to do it.
Since $f_0(0)=0$, then there is an interval $[0,epsilon)$ around zero on which $|f_0|<1/2$. Then $|f_0(epsilon)|leq1/2$.
In order to guarantee that the function $f$ we will construct will satisfy $Vert f-f_0Vert<1$, we just make $f$ "a little positive" when $f_0=1$ and "a little negative" when $f_0=-1$. So choose a small, but positive, $delta$ (how small will be specified later). Let $n:[-1,1]to[-delta,delta]$ be given by
$$n(x)=begincases
x+1-delta&text if -1leq xleq -1+delta\
0&text if -1+deltaleq xleq 1-delta\
x-1+delta&text if 1-deltaleq xleq 1endcases$$
This function $n$ is "a little positive around $1$" and "a little negative" around $-1$. Anyway, define $f(x)=n(f_0(x))$ for all $xin[epsilon,1]$. You can then prove that $|f(x)-f_0(x)|leq 1-delta$ for all $xin[epsilon,1]$.
As long as $delta<1/2$, we also have $f(epsilon)=0$. I'll assume this from now on, because $delta$ can be initially taken very small.
Now we need to define $f$ on $[0,epsilon]$, in a way that $fin Y$. We will define $f$ as a "spike". If we choose $cinmathbbR$, define $f:[0,epsilon]tomathbbR$ in such a way that:
- $f$ is linear on $[0,epsilon/2]$;
- $f$ is linear on $[epsilon/2,epsilon]$
- $f(epsilon/2)=c$.
Then the integral $int_0^epsilon f(x)dx$ is simply the area of the triangle of basis $epsilon$ and heigth $c$, so it is equal to $cepsilon/2$.
We need to guarantee that $int_0^1f(x)dx=0$. So we want that
beginalign*
int_0^epsilon f(x)dx+int_epsilon^1f(x)dx&=0\
epsilon c/2=-int_epsilon^1f(x)dx
endalign*
thus we choose
$$c=-2int_epsilon^1f(x)dx/epsilon$$
Note that
$$|c|leq 2int_epsilon^1|f(x)|dxleq 2delta/epsilon$$
because of the definition of $f=ncirc f_0$ on $[epsilon,1]$, and $|n|leqdelta$. Just choose $delta$ small enough (depending only on $epsilon$) in such a way that $|c|<1/4$.
Repeating all we did so far: as long as we take $delta$ small enough (namely, $delta<1/2$ and $delta<epsilon/8$), we have defined $f$ on all of $[0,1]$, in such a way that
- $fin Y$ (this is precisely why we take $c$ as above).
- for all $xin[epsilon,1]$, $|f(x)-f_0(x)|leq 1-delta$ (this is a computation we did above);
- for all $xin[0,epsilon]$,
$$|f(x)-f_0(x)|leq |f(x)|+|f_0(x)|leq 1/4+1/2=3/4$$
(this is the choice of $epsilon$ and $c$.)
The second and third items above imply that $Vert f-f_0Vertleq max(1-delta,3/4)$, which is smaller than $1$.
answered Aug 6 at 4:33
Luiz Cordeiro
12.2k1041
add a comment |Â
up vote
1
down vote
You can prove that if $Vert f_0Vert=1$, then there is $fin Y$ with $Vert f_0-fVert<1$. Below is a way to do it.
Since $f_0(0)=0$, then there is an interval $[0,epsilon)$ around zero on which $|f_0|<1/2$. Then $|f_0(epsilon)|leq1/2$.
In order to guarantee that the function $f$ we will construct will satisfy $Vert f-f_0Vert<1$, we just make $f$ "a little positive" when $f_0=1$ and "a little negative" when $f_0=-1$. So choose a small, but positive, $delta$ (how small will be specified later). Let $n:[-1,1]to[-delta,delta]$ be given by
$$n(x)=begincases
x+1-delta&text if -1leq xleq -1+delta\
0&text if -1+deltaleq xleq 1-delta\
x-1+delta&text if 1-deltaleq xleq 1endcases$$
This function $n$ is "a little positive around $1$" and "a little negative" around $-1$. Anyway, define $f(x)=n(f_0(x))$ for all $xin[epsilon,1]$. You can then prove that $|f(x)-f_0(x)|leq 1-delta$ for all $xin[epsilon,1]$.
As long as $delta<1/2$, we also have $f(epsilon)=0$. I'll assume this from now on, because $delta$ can be initially taken very small.
Now we need to define $f$ on $[0,epsilon]$, in a way that $fin Y$. We will define $f$ as a "spike". If we choose $cinmathbbR$, define $f:[0,epsilon]tomathbbR$ in such a way that:
- $f$ is linear on $[0,epsilon/2]$;
- $f$ is linear on $[epsilon/2,epsilon]$
- $f(epsilon/2)=c$.
Then the integral $int_0^epsilon f(x)dx$ is simply the area of the triangle of basis $epsilon$ and heigth $c$, so it is equal to $cepsilon/2$.
We need to guarantee that $int_0^1f(x)dx=0$. So we want that
beginalign*
int_0^epsilon f(x)dx+int_epsilon^1f(x)dx&=0\
epsilon c/2=-int_epsilon^1f(x)dx
endalign*
thus we choose
$$c=-2int_epsilon^1f(x)dx/epsilon$$
Note that
$$|c|leq 2int_epsilon^1|f(x)|dxleq 2delta/epsilon$$
because of the definition of $f=ncirc f_0$ on $[epsilon,1]$, and $|n|leqdelta$. Just choose $delta$ small enough (depending only on $epsilon$) in such a way that $|c|<1/4$.
Repeating all we did so far: as long as we take $delta$ small enough (namely, $delta<1/2$ and $delta<epsilon/8$), we have defined $f$ on all of $[0,1]$, in such a way that
- $fin Y$ (this is precisely why we take $c$ as above).
- for all $xin[epsilon,1]$, $|f(x)-f_0(x)|leq 1-delta$ (this is a computation we did above);
- for all $xin[0,epsilon]$,
$$|f(x)-f_0(x)|leq |f(x)|+|f_0(x)|leq 1/4+1/2=3/4$$
(this is the choice of $epsilon$ and $c$.)
The second and third items above imply that $Vert f-f_0Vertleq max(1-delta,3/4)$, which is smaller than $1$.
answered Aug 6 at 4:33
Luiz Cordeiro
12.2k1041
add a comment |Â
up vote
1
down vote
up vote
1
down vote
You can prove that if $Vert f_0Vert=1$, then there is $fin Y$ with $Vert f_0-fVert<1$. Below is a way to do it.
Since $f_0(0)=0$, then there is an interval $[0,epsilon)$ around zero on which $|f_0|<1/2$. Then $|f_0(epsilon)|leq1/2$.
In order to guarantee that the function $f$ we will construct will satisfy $Vert f-f_0Vert<1$, we just make $f$ "a little positive" when $f_0=1$ and "a little negative" when $f_0=-1$. So choose a small, but positive, $delta$ (how small will be specified later). Let $n:[-1,1]to[-delta,delta]$ be given by
$$n(x)=begincases
x+1-delta&text if -1leq xleq -1+delta\
0&text if -1+deltaleq xleq 1-delta\
x-1+delta&text if 1-deltaleq xleq 1endcases$$
This function $n$ is "a little positive around $1$" and "a little negative" around $-1$. Anyway, define $f(x)=n(f_0(x))$ for all $xin[epsilon,1]$. You can then prove that $|f(x)-f_0(x)|leq 1-delta$ for all $xin[epsilon,1]$.
As long as $delta<1/2$, we also have $f(epsilon)=0$. I'll assume this from now on, because $delta$ can be initially taken very small.
Now we need to define $f$ on $[0,epsilon]$, in a way that $fin Y$. We will define $f$ as a "spike". If we choose $cinmathbbR$, define $f:[0,epsilon]tomathbbR$ in such a way that:
- $f$ is linear on $[0,epsilon/2]$;
- $f$ is linear on $[epsilon/2,epsilon]$
- $f(epsilon/2)=c$.
Then the integral $int_0^epsilon f(x)dx$ is simply the area of the triangle of basis $epsilon$ and heigth $c$, so it is equal to $cepsilon/2$.
We need to guarantee that $int_0^1f(x)dx=0$. So we want that
beginalign*
int_0^epsilon f(x)dx+int_epsilon^1f(x)dx&=0\
epsilon c/2=-int_epsilon^1f(x)dx
endalign*
thus we choose
$$c=-2int_epsilon^1f(x)dx/epsilon$$
Note that
$$|c|leq 2int_epsilon^1|f(x)|dxleq 2delta/epsilon$$
because of the definition of $f=ncirc f_0$ on $[epsilon,1]$, and $|n|leqdelta$. Just choose $delta$ small enough (depending only on $epsilon$) in such a way that $|c|<1/4$.
Repeating all we did so far: as long as we take $delta$ small enough (namely, $delta<1/2$ and $delta<epsilon/8$), we have defined $f$ on all of $[0,1]$, in such a way that
- $fin Y$ (this is precisely why we take $c$ as above).
- for all $xin[epsilon,1]$, $|f(x)-f_0(x)|leq 1-delta$ (this is a computation we did above);
- for all $xin[0,epsilon]$,
$$|f(x)-f_0(x)|leq |f(x)|+|f_0(x)|leq 1/4+1/2=3/4$$
(this is the choice of $epsilon$ and $c$.)
The second and third items above imply that $Vert f-f_0Vertleq max(1-delta,3/4)$, which is smaller than $1$.
answered Aug 6 at 4:33
Luiz Cordeiro
12.2k1041
You can prove that if $Vert f_0Vert=1$, then there is $fin Y$ with $Vert f_0-fVert<1$. Below is a way to do it.
Since $f_0(0)=0$, then there is an interval $[0,epsilon)$ around zero on which $|f_0|<1/2$. Then $|f_0(epsilon)|leq1/2$.
In order to guarantee that the function $f$ we will construct will satisfy $Vert f-f_0Vert<1$, we just make $f$ "a little positive" when $f_0=1$ and "a little negative" when $f_0=-1$. So choose a small, but positive, $delta$ (how small will be specified later). Let $n:[-1,1]to[-delta,delta]$ be given by
$$n(x)=begincases
x+1-delta&text if -1leq xleq -1+delta\
0&text if -1+deltaleq xleq 1-delta\
x-1+delta&text if 1-deltaleq xleq 1endcases$$
This function $n$ is "a little positive around $1$" and "a little negative" around $-1$. Anyway, define $f(x)=n(f_0(x))$ for all $xin[epsilon,1]$. You can then prove that $|f(x)-f_0(x)|leq 1-delta$ for all $xin[epsilon,1]$.
As long as $delta<1/2$, we also have $f(epsilon)=0$. I'll assume this from now on, because $delta$ can be initially taken very small.
Now we need to define $f$ on $[0,epsilon]$, in a way that $fin Y$. We will define $f$ as a "spike". If we choose $cinmathbbR$, define $f:[0,epsilon]tomathbbR$ in such a way that:
- $f$ is linear on $[0,epsilon/2]$;
- $f$ is linear on $[epsilon/2,epsilon]$
- $f(epsilon/2)=c$.
Then the integral $int_0^epsilon f(x)dx$ is simply the area of the triangle of basis $epsilon$ and heigth $c$, so it is equal to $cepsilon/2$.
We need to guarantee that $int_0^1f(x)dx=0$. So we want that
beginalign*
int_0^epsilon f(x)dx+int_epsilon^1f(x)dx&=0\
epsilon c/2=-int_epsilon^1f(x)dx
endalign*
thus we choose
$$c=-2int_epsilon^1f(x)dx/epsilon$$
Note that
$$|c|leq 2int_epsilon^1|f(x)|dxleq 2delta/epsilon$$
because of the definition of $f=ncirc f_0$ on $[epsilon,1]$, and $|n|leqdelta$. Just choose $delta$ small enough (depending only on $epsilon$) in such a way that $|c|<1/4$.
Repeating all we did so far: as long as we take $delta$ small enough (namely, $delta<1/2$ and $delta<epsilon/8$), we have defined $f$ on all of $[0,1]$, in such a way that
- $fin Y$ (this is precisely why we take $c$ as above).
- for all $xin[epsilon,1]$, $|f(x)-f_0(x)|leq 1-delta$ (this is a computation we did above);
- for all $xin[0,epsilon]$,
$$|f(x)-f_0(x)|leq |f(x)|+|f_0(x)|leq 1/4+1/2=3/4$$
(this is the choice of $epsilon$ and $c$.)
The second and third items above imply that $Vert f-f_0Vertleq max(1-delta,3/4)$, which is smaller than $1$.
answered Aug 6 at 4:33
Luiz Cordeiro
12.2k1041
edited Aug 6 at 4:43
answered Aug 6 at 4:33
Luiz Cordeiro
12.2k1041
answered Aug 6 at 4:33
Luiz Cordeiro
12.2k1041
answered Aug 6 at 4:33
Luiz Cordeiro
12.2k1041
12.2k1041
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Note that since $0 in Y$, we have $textdist(f_0,Y) le |f_0-0| = 1$.
– copper.hat
Aug 6 at 4:24