Proving the set of polynomials is dense in $C^{1}[0,1]$












1












$begingroup$


Let $C^{1}[0,1]$ denote the set of once-continuously differentiable functions $f:[0,1]rightarrowmathbb R$. For every $fin C^1[0,1]$, find a polynomial $p$ such that $sup_{xin[0,1] }|f(x)-p(x)|<epsilon$ and $sup_{xin[0,1] }|f'(x)-p'(x)|<epsilon$.



So, I can find different polynomials for both, by the Stone-Weirstrass theorem. What I am missing is how to connect them: how can I make sure that the polynomial $p$ I find, or a polynomial $p'$ that I find for the second condition, satisfy the other condition? It really feels like there should be a simple trick to create a third function from some other information.






real-analysis harmonic-analysis







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$endgroup$












  • $begingroup$
    This is essentialy the proof of Stone-Weierstrass theorem using Bernstein Polynoms. Since you are woring on $C^1$ instead of $C^0$, there may be a simpler way.
    $endgroup$
    – nicomezi
    Dec 20 '18 at 12:10












  • $begingroup$
    The thing is, we didn't constructively prove the theorem, meaning we haven't even defined Bernstein polynomials.
    $endgroup$
    – Uri George Peterzil
    Dec 20 '18 at 12:15










  • $begingroup$
    Not sure I understand your remark. Bernstein Polynoms are used to prove the theorem constructively so you do not need the theorem to define them.
    $endgroup$
    – nicomezi
    Dec 20 '18 at 12:16
















1












$begingroup$


Let $C^{1}[0,1]$ denote the set of once-continuously differentiable functions $f:[0,1]rightarrowmathbb R$. For every $fin C^1[0,1]$, find a polynomial $p$ such that $sup_{xin[0,1] }|f(x)-p(x)|<epsilon$ and $sup_{xin[0,1] }|f'(x)-p'(x)|<epsilon$.



So, I can find different polynomials for both, by the Stone-Weirstrass theorem. What I am missing is how to connect them: how can I make sure that the polynomial $p$ I find, or a polynomial $p'$ that I find for the second condition, satisfy the other condition? It really feels like there should be a simple trick to create a third function from some other information.






real-analysis harmonic-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    This is essentialy the proof of Stone-Weierstrass theorem using Bernstein Polynoms. Since you are woring on $C^1$ instead of $C^0$, there may be a simpler way.
    $endgroup$
    – nicomezi
    Dec 20 '18 at 12:10












  • $begingroup$
    The thing is, we didn't constructively prove the theorem, meaning we haven't even defined Bernstein polynomials.
    $endgroup$
    – Uri George Peterzil
    Dec 20 '18 at 12:15










  • $begingroup$
    Not sure I understand your remark. Bernstein Polynoms are used to prove the theorem constructively so you do not need the theorem to define them.
    $endgroup$
    – nicomezi
    Dec 20 '18 at 12:16














1












1








1





$begingroup$


Let $C^{1}[0,1]$ denote the set of once-continuously differentiable functions $f:[0,1]rightarrowmathbb R$. For every $fin C^1[0,1]$, find a polynomial $p$ such that $sup_{xin[0,1] }|f(x)-p(x)|<epsilon$ and $sup_{xin[0,1] }|f'(x)-p'(x)|<epsilon$.



So, I can find different polynomials for both, by the Stone-Weirstrass theorem. What I am missing is how to connect them: how can I make sure that the polynomial $p$ I find, or a polynomial $p'$ that I find for the second condition, satisfy the other condition? It really feels like there should be a simple trick to create a third function from some other information.






real-analysis harmonic-analysis







share|cite|improve this question











$endgroup$




Let $C^{1}[0,1]$ denote the set of once-continuously differentiable functions $f:[0,1]rightarrowmathbb R$. For every $fin C^1[0,1]$, find a polynomial $p$ such that $sup_{xin[0,1] }|f(x)-p(x)|<epsilon$ and $sup_{xin[0,1] }|f'(x)-p'(x)|<epsilon$.



So, I can find different polynomials for both, by the Stone-Weirstrass theorem. What I am missing is how to connect them: how can I make sure that the polynomial $p$ I find, or a polynomial $p'$ that I find for the second condition, satisfy the other condition? It really feels like there should be a simple trick to create a third function from some other information.






real-analysis harmonic-analysis




real-analysis harmonic-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 20 '18 at 12:43









Song

13.8k633




13.8k633










asked Dec 20 '18 at 12:06









Uri George PeterzilUri George Peterzil

949




949












  • $begingroup$
    This is essentialy the proof of Stone-Weierstrass theorem using Bernstein Polynoms. Since you are woring on $C^1$ instead of $C^0$, there may be a simpler way.
    $endgroup$
    – nicomezi
    Dec 20 '18 at 12:10












  • $begingroup$
    The thing is, we didn't constructively prove the theorem, meaning we haven't even defined Bernstein polynomials.
    $endgroup$
    – Uri George Peterzil
    Dec 20 '18 at 12:15










  • $begingroup$
    Not sure I understand your remark. Bernstein Polynoms are used to prove the theorem constructively so you do not need the theorem to define them.
    $endgroup$
    – nicomezi
    Dec 20 '18 at 12:16


















  • $begingroup$
    This is essentialy the proof of Stone-Weierstrass theorem using Bernstein Polynoms. Since you are woring on $C^1$ instead of $C^0$, there may be a simpler way.
    $endgroup$
    – nicomezi
    Dec 20 '18 at 12:10












  • $begingroup$
    The thing is, we didn't constructively prove the theorem, meaning we haven't even defined Bernstein polynomials.
    $endgroup$
    – Uri George Peterzil
    Dec 20 '18 at 12:15










  • $begingroup$
    Not sure I understand your remark. Bernstein Polynoms are used to prove the theorem constructively so you do not need the theorem to define them.
    $endgroup$
    – nicomezi
    Dec 20 '18 at 12:16
















$begingroup$
This is essentialy the proof of Stone-Weierstrass theorem using Bernstein Polynoms. Since you are woring on $C^1$ instead of $C^0$, there may be a simpler way.
$endgroup$
– nicomezi
Dec 20 '18 at 12:10






$begingroup$
This is essentialy the proof of Stone-Weierstrass theorem using Bernstein Polynoms. Since you are woring on $C^1$ instead of $C^0$, there may be a simpler way.
$endgroup$
– nicomezi
Dec 20 '18 at 12:10














$begingroup$
The thing is, we didn't constructively prove the theorem, meaning we haven't even defined Bernstein polynomials.
$endgroup$
– Uri George Peterzil
Dec 20 '18 at 12:15




$begingroup$
The thing is, we didn't constructively prove the theorem, meaning we haven't even defined Bernstein polynomials.
$endgroup$
– Uri George Peterzil
Dec 20 '18 at 12:15












$begingroup$
Not sure I understand your remark. Bernstein Polynoms are used to prove the theorem constructively so you do not need the theorem to define them.
$endgroup$
– nicomezi
Dec 20 '18 at 12:16




$begingroup$
Not sure I understand your remark. Bernstein Polynoms are used to prove the theorem constructively so you do not need the theorem to define them.
$endgroup$
– nicomezi
Dec 20 '18 at 12:16










1 Answer
1






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5












$begingroup$

Since we know that $fin C^1[0,1]$, if follows that $f'in C[0,1]$ and we can find a polynomial $tilde p$ such that $sup_{xin [0,1]} |f'(x)-tilde p(x)| le varepsilon$.



Let $p$ be the antiderivative of $tilde p$ such that $p(0)=f(0)$, i.e.
$$
p(x):=f(0) + int_0^xtilde p(t), dt.
$$
It is obvious that $p$ is a polynomial and that $sup_{xin [0,1]} |f'(x)-p'(x)| le varepsilon$.



Now, we have
$$begin{align}
f(x)-p(x) &= left(f(0) + int_0^x f'(t), dt right)-left(f(0) + int_0^x p'(t), dt right) \
&=int_0^x f'(t)-p'(t), dt,
end{align}$$
hence we can take absolute value and get
$$begin{align}
|f(x)-p(x)| &leint_0^x |f'(t)-p'(t)|, dt \
&le varepsilon int_0^x 1 ,dt \
&le varepsilon
end{align}$$
for all $xin [0,1]$. This proves what you want.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    It seems I misunderstood the problem. Good answer.
    $endgroup$
    – nicomezi
    Dec 20 '18 at 12:21











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1 Answer
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1 Answer
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5












$begingroup$

Since we know that $fin C^1[0,1]$, if follows that $f'in C[0,1]$ and we can find a polynomial $tilde p$ such that $sup_{xin [0,1]} |f'(x)-tilde p(x)| le varepsilon$.



Let $p$ be the antiderivative of $tilde p$ such that $p(0)=f(0)$, i.e.
$$
p(x):=f(0) + int_0^xtilde p(t), dt.
$$
It is obvious that $p$ is a polynomial and that $sup_{xin [0,1]} |f'(x)-p'(x)| le varepsilon$.



Now, we have
$$begin{align}
f(x)-p(x) &= left(f(0) + int_0^x f'(t), dt right)-left(f(0) + int_0^x p'(t), dt right) \
&=int_0^x f'(t)-p'(t), dt,
end{align}$$
hence we can take absolute value and get
$$begin{align}
|f(x)-p(x)| &leint_0^x |f'(t)-p'(t)|, dt \
&le varepsilon int_0^x 1 ,dt \
&le varepsilon
end{align}$$
for all $xin [0,1]$. This proves what you want.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    It seems I misunderstood the problem. Good answer.
    $endgroup$
    – nicomezi
    Dec 20 '18 at 12:21
















5












$begingroup$

Since we know that $fin C^1[0,1]$, if follows that $f'in C[0,1]$ and we can find a polynomial $tilde p$ such that $sup_{xin [0,1]} |f'(x)-tilde p(x)| le varepsilon$.



Let $p$ be the antiderivative of $tilde p$ such that $p(0)=f(0)$, i.e.
$$
p(x):=f(0) + int_0^xtilde p(t), dt.
$$
It is obvious that $p$ is a polynomial and that $sup_{xin [0,1]} |f'(x)-p'(x)| le varepsilon$.



Now, we have
$$begin{align}
f(x)-p(x) &= left(f(0) + int_0^x f'(t), dt right)-left(f(0) + int_0^x p'(t), dt right) \
&=int_0^x f'(t)-p'(t), dt,
end{align}$$
hence we can take absolute value and get
$$begin{align}
|f(x)-p(x)| &leint_0^x |f'(t)-p'(t)|, dt \
&le varepsilon int_0^x 1 ,dt \
&le varepsilon
end{align}$$
for all $xin [0,1]$. This proves what you want.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    It seems I misunderstood the problem. Good answer.
    $endgroup$
    – nicomezi
    Dec 20 '18 at 12:21














5












5








5





$begingroup$

Since we know that $fin C^1[0,1]$, if follows that $f'in C[0,1]$ and we can find a polynomial $tilde p$ such that $sup_{xin [0,1]} |f'(x)-tilde p(x)| le varepsilon$.



Let $p$ be the antiderivative of $tilde p$ such that $p(0)=f(0)$, i.e.
$$
p(x):=f(0) + int_0^xtilde p(t), dt.
$$
It is obvious that $p$ is a polynomial and that $sup_{xin [0,1]} |f'(x)-p'(x)| le varepsilon$.



Now, we have
$$begin{align}
f(x)-p(x) &= left(f(0) + int_0^x f'(t), dt right)-left(f(0) + int_0^x p'(t), dt right) \
&=int_0^x f'(t)-p'(t), dt,
end{align}$$
hence we can take absolute value and get
$$begin{align}
|f(x)-p(x)| &leint_0^x |f'(t)-p'(t)|, dt \
&le varepsilon int_0^x 1 ,dt \
&le varepsilon
end{align}$$
for all $xin [0,1]$. This proves what you want.






share|cite|improve this answer











$endgroup$



Since we know that $fin C^1[0,1]$, if follows that $f'in C[0,1]$ and we can find a polynomial $tilde p$ such that $sup_{xin [0,1]} |f'(x)-tilde p(x)| le varepsilon$.



Let $p$ be the antiderivative of $tilde p$ such that $p(0)=f(0)$, i.e.
$$
p(x):=f(0) + int_0^xtilde p(t), dt.
$$
It is obvious that $p$ is a polynomial and that $sup_{xin [0,1]} |f'(x)-p'(x)| le varepsilon$.



Now, we have
$$begin{align}
f(x)-p(x) &= left(f(0) + int_0^x f'(t), dt right)-left(f(0) + int_0^x p'(t), dt right) \
&=int_0^x f'(t)-p'(t), dt,
end{align}$$
hence we can take absolute value and get
$$begin{align}
|f(x)-p(x)| &leint_0^x |f'(t)-p'(t)|, dt \
&le varepsilon int_0^x 1 ,dt \
&le varepsilon
end{align}$$
for all $xin [0,1]$. This proves what you want.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 20 '18 at 12:44

























answered Dec 20 '18 at 12:20









BigbearZzzBigbearZzz

8,74621652




8,74621652








  • 1




    $begingroup$
    It seems I misunderstood the problem. Good answer.
    $endgroup$
    – nicomezi
    Dec 20 '18 at 12:21














  • 1




    $begingroup$
    It seems I misunderstood the problem. Good answer.
    $endgroup$
    – nicomezi
    Dec 20 '18 at 12:21








1




1




$begingroup$
It seems I misunderstood the problem. Good answer.
$endgroup$
– nicomezi
Dec 20 '18 at 12:21




$begingroup$
It seems I misunderstood the problem. Good answer.
$endgroup$
– nicomezi
Dec 20 '18 at 12:21


















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