Show that for every $epsilon>0$ there exists a polynomial $p(x)$ with the property that…
Let $fin C^1([0, 1])$, with norm $|f|_{C^1}=|f|_infty+|f'|_infty$.
Show that for every $epsilon>0$ there exists a polynomial $p(x)$ with the property that $|f-p|_{C^1}<epsilon$.
My attempt:
Given $f'in C[0, 1]$ and $epsilon>0$, by the Weierstrass approximation theorem, there is a polynomial $q$ such that $displaystyle|f'-q|_infty<frac{epsilon}{2}$.
Since $f'$ is continuous on $[0, 1]$ and $f$ is the indefinite integral of $f'$ on $[0, 1]$, by the fundamental theorem of calculus, $int_0^1 f'(x)dx=f(1)-f(0)$.
Am I on the right track? How do I proceed?
real-analysis
asked Nov 24 at 19:34
Thomas
701415
add a comment |
Let $fin C^1([0, 1])$, with norm $|f|_{C^1}=|f|_infty+|f'|_infty$.
Show that for every $epsilon>0$ there exists a polynomial $p(x)$ with the property that $|f-p|_{C^1}<epsilon$.
My attempt:
Given $f'in C[0, 1]$ and $epsilon>0$, by the Weierstrass approximation theorem, there is a polynomial $q$ such that $displaystyle|f'-q|_infty<frac{epsilon}{2}$.
Since $f'$ is continuous on $[0, 1]$ and $f$ is the indefinite integral of $f'$ on $[0, 1]$, by the fundamental theorem of calculus, $int_0^1 f'(x)dx=f(1)-f(0)$.
Am I on the right track? How do I proceed?
real-analysis
asked Nov 24 at 19:34
Thomas
701415
2
Let $Q$ be an anti-derivative of $q$ so that $Q(0)=f(0)$. This is again a polynomial. Note now that $$|Q(x)-f(x)|=left|int_0^x (q(t)-f'(t))dtright|≤int_0^x |q-f'|_infty,dt ≤|q-f'|_infty≤epsilon/2.$$ Then $|Q-f|_{C^1}≤epsilon$ follows.
– s.harp
Nov 24 at 19:43
add a comment |
Let $fin C^1([0, 1])$, with norm $|f|_{C^1}=|f|_infty+|f'|_infty$.
Show that for every $epsilon>0$ there exists a polynomial $p(x)$ with the property that $|f-p|_{C^1}<epsilon$.
My attempt:
Given $f'in C[0, 1]$ and $epsilon>0$, by the Weierstrass approximation theorem, there is a polynomial $q$ such that $displaystyle|f'-q|_infty<frac{epsilon}{2}$.
Since $f'$ is continuous on $[0, 1]$ and $f$ is the indefinite integral of $f'$ on $[0, 1]$, by the fundamental theorem of calculus, $int_0^1 f'(x)dx=f(1)-f(0)$.
Am I on the right track? How do I proceed?
real-analysis
asked Nov 24 at 19:34
Thomas
701415
Let $fin C^1([0, 1])$, with norm $|f|_{C^1}=|f|_infty+|f'|_infty$.
Show that for every $epsilon>0$ there exists a polynomial $p(x)$ with the property that $|f-p|_{C^1}<epsilon$.
My attempt:
Given $f'in C[0, 1]$ and $epsilon>0$, by the Weierstrass approximation theorem, there is a polynomial $q$ such that $displaystyle|f'-q|_infty<frac{epsilon}{2}$.
Since $f'$ is continuous on $[0, 1]$ and $f$ is the indefinite integral of $f'$ on $[0, 1]$, by the fundamental theorem of calculus, $int_0^1 f'(x)dx=f(1)-f(0)$.
Am I on the right track? How do I proceed?
real-analysis
real-analysis
asked Nov 24 at 19:34
Thomas
701415
asked Nov 24 at 19:34
Thomas
701415
asked Nov 24 at 19:34
Thomas
701415
asked Nov 24 at 19:34
Thomas
701415
asked Nov 24 at 19:34
Thomas
701415
701415
2
Let $Q$ be an anti-derivative of $q$ so that $Q(0)=f(0)$. This is again a polynomial. Note now that $$|Q(x)-f(x)|=left|int_0^x (q(t)-f'(t))dtright|≤int_0^x |q-f'|_infty,dt ≤|q-f'|_infty≤epsilon/2.$$ Then $|Q-f|_{C^1}≤epsilon$ follows.
– s.harp
Nov 24 at 19:43
add a comment |
2
Let $Q$ be an anti-derivative of $q$ so that $Q(0)=f(0)$. This is again a polynomial. Note now that $$|Q(x)-f(x)|=left|int_0^x (q(t)-f'(t))dtright|≤int_0^x |q-f'|_infty,dt ≤|q-f'|_infty≤epsilon/2.$$ Then $|Q-f|_{C^1}≤epsilon$ follows.
– s.harp
Nov 24 at 19:43
2
2
Let $Q$ be an anti-derivative of $q$ so that $Q(0)=f(0)$. This is again a polynomial. Note now that $$|Q(x)-f(x)|=left|int_0^x (q(t)-f'(t))dtright|≤int_0^x |q-f'|_infty,dt ≤|q-f'|_infty≤epsilon/2.$$ Then $|Q-f|_{C^1}≤epsilon$ follows.
– s.harp
Nov 24 at 19:43
Let $Q$ be an anti-derivative of $q$ so that $Q(0)=f(0)$. This is again a polynomial. Note now that $$|Q(x)-f(x)|=left|int_0^x (q(t)-f'(t))dtright|≤int_0^x |q-f'|_infty,dt ≤|q-f'|_infty≤epsilon/2.$$ Then $|Q-f|_{C^1}≤epsilon$ follows.
– s.harp
Nov 24 at 19:43
add a comment |
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2
Let $Q$ be an anti-derivative of $q$ so that $Q(0)=f(0)$. This is again a polynomial. Note now that $$|Q(x)-f(x)|=left|int_0^x (q(t)-f'(t))dtright|≤int_0^x |q-f'|_infty,dt ≤|q-f'|_infty≤epsilon/2.$$ Then $|Q-f|_{C^1}≤epsilon$ follows.
– s.harp
Nov 24 at 19:43