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Approximation by smooth functions by changing values at arbitrarily small interval
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Assume $fin C[0,1]$ is smooth (i.e. infinitely many times differentiable) on $(0,frac 12)$ and $(frac 12,1)$. Let $epsilon>0$ be arbitrarily small.
Can we approximate $f$ in supremum norm by functions $f_n$ smooth on $(0,1)$
and such that $f_n(x)=f(x)$ for $xin[0,frac 12-epsilon)cup(frac 12+epsilon,1]$? Is there a construction of such approximation?
real-analysis approximation weierstrass-approximation
asked Jul 19 at 19:53
lucas
111
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Assume $fin C[0,1]$ is smooth (i.e. infinitely many times differentiable) on $(0,frac 12)$ and $(frac 12,1)$. Let $epsilon>0$ be arbitrarily small.
Can we approximate $f$ in supremum norm by functions $f_n$ smooth on $(0,1)$
and such that $f_n(x)=f(x)$ for $xin[0,frac 12-epsilon)cup(frac 12+epsilon,1]$? Is there a construction of such approximation?
real-analysis approximation weierstrass-approximation
asked Jul 19 at 19:53
lucas
111
On interval $[0,1]$.
– lucas
Jul 19 at 20:02
I guess you can. Let $f_n$ be constantly $f(1/2)$ on $(1/2-epsilon/2,1/2+epsilon/2)$ and connect the rest smoothly.
– amsmath
Jul 19 at 20:03
Check out smooth bump functions or smooth step functions.
– user357980
Jul 20 at 0:10
Provided that f(x) is continuous at x=1/2 Otherwise, no.
– DanielWainfleet
Jul 20 at 20:20
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Assume $fin C[0,1]$ is smooth (i.e. infinitely many times differentiable) on $(0,frac 12)$ and $(frac 12,1)$. Let $epsilon>0$ be arbitrarily small.
Can we approximate $f$ in supremum norm by functions $f_n$ smooth on $(0,1)$
and such that $f_n(x)=f(x)$ for $xin[0,frac 12-epsilon)cup(frac 12+epsilon,1]$? Is there a construction of such approximation?
real-analysis approximation weierstrass-approximation
asked Jul 19 at 19:53
lucas
111
Assume $fin C[0,1]$ is smooth (i.e. infinitely many times differentiable) on $(0,frac 12)$ and $(frac 12,1)$. Let $epsilon>0$ be arbitrarily small.
Can we approximate $f$ in supremum norm by functions $f_n$ smooth on $(0,1)$
and such that $f_n(x)=f(x)$ for $xin[0,frac 12-epsilon)cup(frac 12+epsilon,1]$? Is there a construction of such approximation?
real-analysis approximation weierstrass-approximation
asked Jul 19 at 19:53
lucas
111
asked Jul 19 at 19:53
lucas
111
asked Jul 19 at 19:53
lucas
111
asked Jul 19 at 19:53
lucas
111
111
On interval $[0,1]$.
– lucas
Jul 19 at 20:02
I guess you can. Let $f_n$ be constantly $f(1/2)$ on $(1/2-epsilon/2,1/2+epsilon/2)$ and connect the rest smoothly.
– amsmath
Jul 19 at 20:03
Check out smooth bump functions or smooth step functions.
– user357980
Jul 20 at 0:10
Provided that f(x) is continuous at x=1/2 Otherwise, no.
– DanielWainfleet
Jul 20 at 20:20
add a comment |Â
On interval $[0,1]$.
– lucas
Jul 19 at 20:02
I guess you can. Let $f_n$ be constantly $f(1/2)$ on $(1/2-epsilon/2,1/2+epsilon/2)$ and connect the rest smoothly.
– amsmath
Jul 19 at 20:03
Check out smooth bump functions or smooth step functions.
– user357980
Jul 20 at 0:10
Provided that f(x) is continuous at x=1/2 Otherwise, no.
– DanielWainfleet
Jul 20 at 20:20
On interval $[0,1]$.
– lucas
Jul 19 at 20:02
On interval $[0,1]$.
– lucas
Jul 19 at 20:02
I guess you can. Let $f_n$ be constantly $f(1/2)$ on $(1/2-epsilon/2,1/2+epsilon/2)$ and connect the rest smoothly.
– amsmath
Jul 19 at 20:03
I guess you can. Let $f_n$ be constantly $f(1/2)$ on $(1/2-epsilon/2,1/2+epsilon/2)$ and connect the rest smoothly.
– amsmath
Jul 19 at 20:03
Check out smooth bump functions or smooth step functions.
– user357980
Jul 20 at 0:10
Check out smooth bump functions or smooth step functions.
– user357980
Jul 20 at 0:10
Provided that f(x) is continuous at x=1/2 Otherwise, no.
– DanielWainfleet
Jul 20 at 20:20
Provided that f(x) is continuous at x=1/2 Otherwise, no.
– DanielWainfleet
Jul 20 at 20:20
add a comment |Â
1 Answer
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It is true that there exists a sequence of smooth functions that converge to $f$ in the sup norm, e.g. the Bernstein polynomial. But your statement is clearly false. Indeed, if $f_n(x) = f(x)$ on $[0,frac12−epsilon)∪(frac12-epsilon,1]$, then we see $f_n$ is differentiable at $x=1/2$ if and only if $f$ is. Indeed, this follows from differentiation is a local property. So if $f$ is not differentiable at $x=1/2$ then $f_n$ cannot be as well, so a counter example would be $f(x)=|x-1/2|$. However, here's a proof of $f_n rightarrow f$ on sup norm even if $f$ is just continuous on $[0,1]$.
This can only possibly be true on subsets of $[0,1]$ as if $f_n rightarrow f$ with the sup norm where $f_n$ are smooth, then they converge uniformly, so $f$ is continuous. Hence, this theorem can possibly only hold on subsets of $[0,1]$ as we do not know the continuity of $f$ outside of $[0,1]$.
Now let $[a,b] subset [0,1]$, then we see $f$ is uniformly continuous on $[a,b]$ in fact any interval will work as $f$ is uniformly continous on $[0,1]$. Now by Weiestrass's Approximation Theorem (https://en.wikipedia.org/wiki/Stone%E2%80%93Weierstrass_theorem#Weierstrass_approximation_theorem)
there exists a sequence of polynomials $p_n$ such that $p_n rightarrow f$ uniformly, hence $sup_x in [0,1]||f-p_n|| rightarrow 0$. In particular, the $p_n$ are the Bernstein polynomials (https://en.wikipedia.org/wiki/Bernstein_polynomial). (I'll let you read wikipedia for the constructive proof)
answered Jul 20 at 5:48
Raymond Chu
1,03719
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
It is true that there exists a sequence of smooth functions that converge to $f$ in the sup norm, e.g. the Bernstein polynomial. But your statement is clearly false. Indeed, if $f_n(x) = f(x)$ on $[0,frac12−epsilon)∪(frac12-epsilon,1]$, then we see $f_n$ is differentiable at $x=1/2$ if and only if $f$ is. Indeed, this follows from differentiation is a local property. So if $f$ is not differentiable at $x=1/2$ then $f_n$ cannot be as well, so a counter example would be $f(x)=|x-1/2|$. However, here's a proof of $f_n rightarrow f$ on sup norm even if $f$ is just continuous on $[0,1]$.
This can only possibly be true on subsets of $[0,1]$ as if $f_n rightarrow f$ with the sup norm where $f_n$ are smooth, then they converge uniformly, so $f$ is continuous. Hence, this theorem can possibly only hold on subsets of $[0,1]$ as we do not know the continuity of $f$ outside of $[0,1]$.
Now let $[a,b] subset [0,1]$, then we see $f$ is uniformly continuous on $[a,b]$ in fact any interval will work as $f$ is uniformly continous on $[0,1]$. Now by Weiestrass's Approximation Theorem (https://en.wikipedia.org/wiki/Stone%E2%80%93Weierstrass_theorem#Weierstrass_approximation_theorem)
there exists a sequence of polynomials $p_n$ such that $p_n rightarrow f$ uniformly, hence $sup_x in [0,1]||f-p_n|| rightarrow 0$. In particular, the $p_n$ are the Bernstein polynomials (https://en.wikipedia.org/wiki/Bernstein_polynomial). (I'll let you read wikipedia for the constructive proof)
answered Jul 20 at 5:48
Raymond Chu
1,03719
add a comment |Â
up vote
2
down vote
It is true that there exists a sequence of smooth functions that converge to $f$ in the sup norm, e.g. the Bernstein polynomial. But your statement is clearly false. Indeed, if $f_n(x) = f(x)$ on $[0,frac12−epsilon)∪(frac12-epsilon,1]$, then we see $f_n$ is differentiable at $x=1/2$ if and only if $f$ is. Indeed, this follows from differentiation is a local property. So if $f$ is not differentiable at $x=1/2$ then $f_n$ cannot be as well, so a counter example would be $f(x)=|x-1/2|$. However, here's a proof of $f_n rightarrow f$ on sup norm even if $f$ is just continuous on $[0,1]$.
This can only possibly be true on subsets of $[0,1]$ as if $f_n rightarrow f$ with the sup norm where $f_n$ are smooth, then they converge uniformly, so $f$ is continuous. Hence, this theorem can possibly only hold on subsets of $[0,1]$ as we do not know the continuity of $f$ outside of $[0,1]$.
Now let $[a,b] subset [0,1]$, then we see $f$ is uniformly continuous on $[a,b]$ in fact any interval will work as $f$ is uniformly continous on $[0,1]$. Now by Weiestrass's Approximation Theorem (https://en.wikipedia.org/wiki/Stone%E2%80%93Weierstrass_theorem#Weierstrass_approximation_theorem)
there exists a sequence of polynomials $p_n$ such that $p_n rightarrow f$ uniformly, hence $sup_x in [0,1]||f-p_n|| rightarrow 0$. In particular, the $p_n$ are the Bernstein polynomials (https://en.wikipedia.org/wiki/Bernstein_polynomial). (I'll let you read wikipedia for the constructive proof)
answered Jul 20 at 5:48
Raymond Chu
1,03719
add a comment |Â
up vote
2
down vote
up vote
2
down vote
It is true that there exists a sequence of smooth functions that converge to $f$ in the sup norm, e.g. the Bernstein polynomial. But your statement is clearly false. Indeed, if $f_n(x) = f(x)$ on $[0,frac12−epsilon)∪(frac12-epsilon,1]$, then we see $f_n$ is differentiable at $x=1/2$ if and only if $f$ is. Indeed, this follows from differentiation is a local property. So if $f$ is not differentiable at $x=1/2$ then $f_n$ cannot be as well, so a counter example would be $f(x)=|x-1/2|$. However, here's a proof of $f_n rightarrow f$ on sup norm even if $f$ is just continuous on $[0,1]$.
This can only possibly be true on subsets of $[0,1]$ as if $f_n rightarrow f$ with the sup norm where $f_n$ are smooth, then they converge uniformly, so $f$ is continuous. Hence, this theorem can possibly only hold on subsets of $[0,1]$ as we do not know the continuity of $f$ outside of $[0,1]$.
Now let $[a,b] subset [0,1]$, then we see $f$ is uniformly continuous on $[a,b]$ in fact any interval will work as $f$ is uniformly continous on $[0,1]$. Now by Weiestrass's Approximation Theorem (https://en.wikipedia.org/wiki/Stone%E2%80%93Weierstrass_theorem#Weierstrass_approximation_theorem)
there exists a sequence of polynomials $p_n$ such that $p_n rightarrow f$ uniformly, hence $sup_x in [0,1]||f-p_n|| rightarrow 0$. In particular, the $p_n$ are the Bernstein polynomials (https://en.wikipedia.org/wiki/Bernstein_polynomial). (I'll let you read wikipedia for the constructive proof)
answered Jul 20 at 5:48
Raymond Chu
1,03719
It is true that there exists a sequence of smooth functions that converge to $f$ in the sup norm, e.g. the Bernstein polynomial. But your statement is clearly false. Indeed, if $f_n(x) = f(x)$ on $[0,frac12−epsilon)∪(frac12-epsilon,1]$, then we see $f_n$ is differentiable at $x=1/2$ if and only if $f$ is. Indeed, this follows from differentiation is a local property. So if $f$ is not differentiable at $x=1/2$ then $f_n$ cannot be as well, so a counter example would be $f(x)=|x-1/2|$. However, here's a proof of $f_n rightarrow f$ on sup norm even if $f$ is just continuous on $[0,1]$.
This can only possibly be true on subsets of $[0,1]$ as if $f_n rightarrow f$ with the sup norm where $f_n$ are smooth, then they converge uniformly, so $f$ is continuous. Hence, this theorem can possibly only hold on subsets of $[0,1]$ as we do not know the continuity of $f$ outside of $[0,1]$.
Now let $[a,b] subset [0,1]$, then we see $f$ is uniformly continuous on $[a,b]$ in fact any interval will work as $f$ is uniformly continous on $[0,1]$. Now by Weiestrass's Approximation Theorem (https://en.wikipedia.org/wiki/Stone%E2%80%93Weierstrass_theorem#Weierstrass_approximation_theorem)
there exists a sequence of polynomials $p_n$ such that $p_n rightarrow f$ uniformly, hence $sup_x in [0,1]||f-p_n|| rightarrow 0$. In particular, the $p_n$ are the Bernstein polynomials (https://en.wikipedia.org/wiki/Bernstein_polynomial). (I'll let you read wikipedia for the constructive proof)
answered Jul 20 at 5:48
Raymond Chu
1,03719
edited Jul 20 at 5:53
answered Jul 20 at 5:48
Raymond Chu
1,03719
answered Jul 20 at 5:48
Raymond Chu
1,03719
answered Jul 20 at 5:48
Raymond Chu
1,03719
1,03719
add a comment |Â
add a comment |Â
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On interval $[0,1]$.
– lucas
Jul 19 at 20:02
I guess you can. Let $f_n$ be constantly $f(1/2)$ on $(1/2-epsilon/2,1/2+epsilon/2)$ and connect the rest smoothly.
– amsmath
Jul 19 at 20:03
Check out smooth bump functions or smooth step functions.
– user357980
Jul 20 at 0:10
Provided that f(x) is continuous at x=1/2 Otherwise, no.
– DanielWainfleet
Jul 20 at 20:20