Is the set ${|f(0)|: int_{0}^{1}|f(t)|dtle1}$ bounded?
Let $x_0 in [0,1]$ and define $T:C[0,1] rightarrow mathbb{R}$ by $T_{x_0}(f)=f(x_0)$. Let $||cdot||_1$ be a norm on $C[0,1]$. Is $T_0$ bounded or not? That is, is the set
$$
left{|T_{0}(f)|:||f||_1 leq 1right}={|f(0)|:||f||_1 leq 1,f in C[0,1]}
$$ bounded? Since $||f||_1:=int_{0}^{1}|f(t)|dt$, the question may be equivalent to the following:
Let $f:[0,1] rightarrow mathbb{R}$ be continuous. Is the set $$left{|f(0)|: int_{0}^{1}|f(t)|dt leq 1right}$$ bounded?
I guess the answer is no. Because, for example, we can have a function whose graph is a narrow spike at the origin but with infinite height. The area enclosed by the graph may be 1 but the value at the origin $f(0)$ which is its height is infinite.
But how can I prove this formally?
calculus real-analysis functional-analysis analysis
edited Nov 29 at 21:45
Jean-Claude Arbaut
14.7k63464
asked Nov 29 at 21:30
bbw
47038
|
show 2 more comments
Let $x_0 in [0,1]$ and define $T:C[0,1] rightarrow mathbb{R}$ by $T_{x_0}(f)=f(x_0)$. Let $||cdot||_1$ be a norm on $C[0,1]$. Is $T_0$ bounded or not? That is, is the set
$$
left{|T_{0}(f)|:||f||_1 leq 1right}={|f(0)|:||f||_1 leq 1,f in C[0,1]}
$$ bounded? Since $||f||_1:=int_{0}^{1}|f(t)|dt$, the question may be equivalent to the following:
Let $f:[0,1] rightarrow mathbb{R}$ be continuous. Is the set $$left{|f(0)|: int_{0}^{1}|f(t)|dt leq 1right}$$ bounded?
I guess the answer is no. Because, for example, we can have a function whose graph is a narrow spike at the origin but with infinite height. The area enclosed by the graph may be 1 but the value at the origin $f(0)$ which is its height is infinite.
But how can I prove this formally?
calculus real-analysis functional-analysis analysis
edited Nov 29 at 21:45
Jean-Claude Arbaut
14.7k63464
asked Nov 29 at 21:30
bbw
47038
As written, this makes little sense. What is the precise definition of the set? There's no condition placed on anything. Is it a set of functions?
– Alex R.
Nov 29 at 21:32
This is not a set... you need to assume something about $int_0^1 | f(t)|dt$. I mean it should be a true/false statement...
– Yanko
Nov 29 at 21:32
Is the RHS trying to say that $f$ is absolutely integrable on $[0,1]$? If so, you have the right idea, consider the function $1/sqrt{x}$ and its integral on $[0,1]$
– rubikscube09
Nov 29 at 21:35
Sorry for the confusion. Just edited the description to make it clearer.
– bbw
Nov 29 at 21:37
1
Is it really a problem? What about a constant coefficient?
– Jean-Claude Arbaut
Nov 29 at 21:51
|
show 2 more comments
Let $x_0 in [0,1]$ and define $T:C[0,1] rightarrow mathbb{R}$ by $T_{x_0}(f)=f(x_0)$. Let $||cdot||_1$ be a norm on $C[0,1]$. Is $T_0$ bounded or not? That is, is the set
$$
left{|T_{0}(f)|:||f||_1 leq 1right}={|f(0)|:||f||_1 leq 1,f in C[0,1]}
$$ bounded? Since $||f||_1:=int_{0}^{1}|f(t)|dt$, the question may be equivalent to the following:
Let $f:[0,1] rightarrow mathbb{R}$ be continuous. Is the set $$left{|f(0)|: int_{0}^{1}|f(t)|dt leq 1right}$$ bounded?
I guess the answer is no. Because, for example, we can have a function whose graph is a narrow spike at the origin but with infinite height. The area enclosed by the graph may be 1 but the value at the origin $f(0)$ which is its height is infinite.
But how can I prove this formally?
calculus real-analysis functional-analysis analysis
edited Nov 29 at 21:45
Jean-Claude Arbaut
14.7k63464
asked Nov 29 at 21:30
bbw
47038
Let $x_0 in [0,1]$ and define $T:C[0,1] rightarrow mathbb{R}$ by $T_{x_0}(f)=f(x_0)$. Let $||cdot||_1$ be a norm on $C[0,1]$. Is $T_0$ bounded or not? That is, is the set
$$
left{|T_{0}(f)|:||f||_1 leq 1right}={|f(0)|:||f||_1 leq 1,f in C[0,1]}
$$ bounded? Since $||f||_1:=int_{0}^{1}|f(t)|dt$, the question may be equivalent to the following:
Let $f:[0,1] rightarrow mathbb{R}$ be continuous. Is the set $$left{|f(0)|: int_{0}^{1}|f(t)|dt leq 1right}$$ bounded?
I guess the answer is no. Because, for example, we can have a function whose graph is a narrow spike at the origin but with infinite height. The area enclosed by the graph may be 1 but the value at the origin $f(0)$ which is its height is infinite.
But how can I prove this formally?
calculus real-analysis functional-analysis analysis
calculus real-analysis functional-analysis analysis
edited Nov 29 at 21:45
Jean-Claude Arbaut
14.7k63464
asked Nov 29 at 21:30
bbw
47038
edited Nov 29 at 21:45
Jean-Claude Arbaut
14.7k63464
asked Nov 29 at 21:30
bbw
47038
edited Nov 29 at 21:45
Jean-Claude Arbaut
14.7k63464
edited Nov 29 at 21:45
Jean-Claude Arbaut
14.7k63464
edited Nov 29 at 21:45
Jean-Claude Arbaut
14.7k63464
14.7k63464
asked Nov 29 at 21:30
bbw
47038
asked Nov 29 at 21:30
bbw
47038
asked Nov 29 at 21:30
bbw
47038
47038
As written, this makes little sense. What is the precise definition of the set? There's no condition placed on anything. Is it a set of functions?
– Alex R.
Nov 29 at 21:32
This is not a set... you need to assume something about $int_0^1 | f(t)|dt$. I mean it should be a true/false statement...
– Yanko
Nov 29 at 21:32
Is the RHS trying to say that $f$ is absolutely integrable on $[0,1]$? If so, you have the right idea, consider the function $1/sqrt{x}$ and its integral on $[0,1]$
– rubikscube09
Nov 29 at 21:35
Sorry for the confusion. Just edited the description to make it clearer.
– bbw
Nov 29 at 21:37
1
Is it really a problem? What about a constant coefficient?
– Jean-Claude Arbaut
Nov 29 at 21:51
|
show 2 more comments
As written, this makes little sense. What is the precise definition of the set? There's no condition placed on anything. Is it a set of functions?
– Alex R.
Nov 29 at 21:32
This is not a set... you need to assume something about $int_0^1 | f(t)|dt$. I mean it should be a true/false statement...
– Yanko
Nov 29 at 21:32
Is the RHS trying to say that $f$ is absolutely integrable on $[0,1]$? If so, you have the right idea, consider the function $1/sqrt{x}$ and its integral on $[0,1]$
– rubikscube09
Nov 29 at 21:35
Sorry for the confusion. Just edited the description to make it clearer.
– bbw
Nov 29 at 21:37
1
Is it really a problem? What about a constant coefficient?
– Jean-Claude Arbaut
Nov 29 at 21:51
As written, this makes little sense. What is the precise definition of the set? There's no condition placed on anything. Is it a set of functions?
– Alex R.
Nov 29 at 21:32
As written, this makes little sense. What is the precise definition of the set? There's no condition placed on anything. Is it a set of functions?
– Alex R.
Nov 29 at 21:32
This is not a set... you need to assume something about $int_0^1 | f(t)|dt$. I mean it should be a true/false statement...
– Yanko
Nov 29 at 21:32
This is not a set... you need to assume something about $int_0^1 | f(t)|dt$. I mean it should be a true/false statement...
– Yanko
Nov 29 at 21:32
Is the RHS trying to say that $f$ is absolutely integrable on $[0,1]$? If so, you have the right idea, consider the function $1/sqrt{x}$ and its integral on $[0,1]$
– rubikscube09
Nov 29 at 21:35
Is the RHS trying to say that $f$ is absolutely integrable on $[0,1]$? If so, you have the right idea, consider the function $1/sqrt{x}$ and its integral on $[0,1]$
– rubikscube09
Nov 29 at 21:35
Sorry for the confusion. Just edited the description to make it clearer.
– bbw
Nov 29 at 21:37
Sorry for the confusion. Just edited the description to make it clearer.
– bbw
Nov 29 at 21:37
1
1
Is it really a problem? What about a constant coefficient?
– Jean-Claude Arbaut
Nov 29 at 21:51
Is it really a problem? What about a constant coefficient?
– Jean-Claude Arbaut
Nov 29 at 21:51
|
show 2 more comments
4 Answers
4
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oldest
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No, it is not bounded. Define, for each $ninmathbb N$,$$begin{array}{rccc}f_ncolon&[0,1]&longrightarrow&mathbb R\&t&mapsto&begin{cases}n-n^2t&text{ if }tleqslantfrac1n\0&text{ otherwise.}end{cases}end{array}$$Then$$int_0^1bigllvert f_n(t)bigrrvert,mathrm dt=frac12,$$but $f_n(0)=n$.
answered Nov 29 at 21:50
José Carlos Santos
149k22117219
add a comment |
We can take a function $f_n$ such that $f_n$ is affine on $(0,1/n)$, $f_n(0)=2n$ and $f_n(1/n)=0$ and $0$ for the other values of the interval. Then $leftlVert f_nrightrVert_1=1$.
For a formal example: define for each positive integer $n$ the function $f_n$ in the following way: $f_n(t)=-2n^2t+2n$ for $0leqslant tleqslant 1/n$ and $f_n(t)=0$ for $1/nlt tleqslant 1$.
answered Nov 29 at 21:51
Davide Giraudo
125k16150259
add a comment |
Consider the function $f_n(x)=2n(1/n-x)1_{[0,1/n]}(x).$ Then $int_0^1 |f_n(x)|dx=1$ but $f_n(0)=2n.$
answered Nov 29 at 21:52
John_Wick
1,356111
add a comment |
For every $a>0$, the function
$$f_a(x)=frac{2 a e^{-a^2 x^2}}{sqrt{pi} textrm{erf}(a)}$$ with the error function $textrm{erf}(a)=frac{2}{pi}int_0^a e^{-t^2}dt$ is in the set ${|f(0)|: int_{0}^{1}|f(t)|dt=1}$ and evaluates to $f_a(0)=frac{2 a }{sqrt{pi} textrm{erf}(a)}$. Because $textrm{erf}(a) rightarrow 1$ as $arightarrow infty$, we obtain $f_a(0)$ arbitrarily large as we increase $a$. As a consequence, $f_a(0)$ and thus your set are unbounded.
For every $a>0$, the function $f_a(x)$ is in $C[0,1]$ and even infinitely differentiable on $[0,1]$.
answered Nov 29 at 21:53
M1183
943
add a comment |
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4 Answers
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active
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4 Answers
4
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oldest
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votes
No, it is not bounded. Define, for each $ninmathbb N$,$$begin{array}{rccc}f_ncolon&[0,1]&longrightarrow&mathbb R\&t&mapsto&begin{cases}n-n^2t&text{ if }tleqslantfrac1n\0&text{ otherwise.}end{cases}end{array}$$Then$$int_0^1bigllvert f_n(t)bigrrvert,mathrm dt=frac12,$$but $f_n(0)=n$.
answered Nov 29 at 21:50
José Carlos Santos
149k22117219
add a comment |
No, it is not bounded. Define, for each $ninmathbb N$,$$begin{array}{rccc}f_ncolon&[0,1]&longrightarrow&mathbb R\&t&mapsto&begin{cases}n-n^2t&text{ if }tleqslantfrac1n\0&text{ otherwise.}end{cases}end{array}$$Then$$int_0^1bigllvert f_n(t)bigrrvert,mathrm dt=frac12,$$but $f_n(0)=n$.
answered Nov 29 at 21:50
José Carlos Santos
149k22117219
add a comment |
No, it is not bounded. Define, for each $ninmathbb N$,$$begin{array}{rccc}f_ncolon&[0,1]&longrightarrow&mathbb R\&t&mapsto&begin{cases}n-n^2t&text{ if }tleqslantfrac1n\0&text{ otherwise.}end{cases}end{array}$$Then$$int_0^1bigllvert f_n(t)bigrrvert,mathrm dt=frac12,$$but $f_n(0)=n$.
answered Nov 29 at 21:50
José Carlos Santos
149k22117219
No, it is not bounded. Define, for each $ninmathbb N$,$$begin{array}{rccc}f_ncolon&[0,1]&longrightarrow&mathbb R\&t&mapsto&begin{cases}n-n^2t&text{ if }tleqslantfrac1n\0&text{ otherwise.}end{cases}end{array}$$Then$$int_0^1bigllvert f_n(t)bigrrvert,mathrm dt=frac12,$$but $f_n(0)=n$.
answered Nov 29 at 21:50
José Carlos Santos
149k22117219
answered Nov 29 at 21:50
José Carlos Santos
149k22117219
answered Nov 29 at 21:50
José Carlos Santos
149k22117219
answered Nov 29 at 21:50
José Carlos Santos
149k22117219
149k22117219
add a comment |
add a comment |
We can take a function $f_n$ such that $f_n$ is affine on $(0,1/n)$, $f_n(0)=2n$ and $f_n(1/n)=0$ and $0$ for the other values of the interval. Then $leftlVert f_nrightrVert_1=1$.
For a formal example: define for each positive integer $n$ the function $f_n$ in the following way: $f_n(t)=-2n^2t+2n$ for $0leqslant tleqslant 1/n$ and $f_n(t)=0$ for $1/nlt tleqslant 1$.
answered Nov 29 at 21:51
Davide Giraudo
125k16150259
add a comment |
We can take a function $f_n$ such that $f_n$ is affine on $(0,1/n)$, $f_n(0)=2n$ and $f_n(1/n)=0$ and $0$ for the other values of the interval. Then $leftlVert f_nrightrVert_1=1$.
For a formal example: define for each positive integer $n$ the function $f_n$ in the following way: $f_n(t)=-2n^2t+2n$ for $0leqslant tleqslant 1/n$ and $f_n(t)=0$ for $1/nlt tleqslant 1$.
answered Nov 29 at 21:51
Davide Giraudo
125k16150259
add a comment |
We can take a function $f_n$ such that $f_n$ is affine on $(0,1/n)$, $f_n(0)=2n$ and $f_n(1/n)=0$ and $0$ for the other values of the interval. Then $leftlVert f_nrightrVert_1=1$.
For a formal example: define for each positive integer $n$ the function $f_n$ in the following way: $f_n(t)=-2n^2t+2n$ for $0leqslant tleqslant 1/n$ and $f_n(t)=0$ for $1/nlt tleqslant 1$.
answered Nov 29 at 21:51
Davide Giraudo
125k16150259
We can take a function $f_n$ such that $f_n$ is affine on $(0,1/n)$, $f_n(0)=2n$ and $f_n(1/n)=0$ and $0$ for the other values of the interval. Then $leftlVert f_nrightrVert_1=1$.
For a formal example: define for each positive integer $n$ the function $f_n$ in the following way: $f_n(t)=-2n^2t+2n$ for $0leqslant tleqslant 1/n$ and $f_n(t)=0$ for $1/nlt tleqslant 1$.
answered Nov 29 at 21:51
Davide Giraudo
125k16150259
answered Nov 29 at 21:51
Davide Giraudo
125k16150259
answered Nov 29 at 21:51
Davide Giraudo
125k16150259
answered Nov 29 at 21:51
Davide Giraudo
125k16150259
125k16150259
add a comment |
add a comment |
Consider the function $f_n(x)=2n(1/n-x)1_{[0,1/n]}(x).$ Then $int_0^1 |f_n(x)|dx=1$ but $f_n(0)=2n.$
answered Nov 29 at 21:52
John_Wick
1,356111
add a comment |
Consider the function $f_n(x)=2n(1/n-x)1_{[0,1/n]}(x).$ Then $int_0^1 |f_n(x)|dx=1$ but $f_n(0)=2n.$
answered Nov 29 at 21:52
John_Wick
1,356111
add a comment |
Consider the function $f_n(x)=2n(1/n-x)1_{[0,1/n]}(x).$ Then $int_0^1 |f_n(x)|dx=1$ but $f_n(0)=2n.$
answered Nov 29 at 21:52
John_Wick
1,356111
Consider the function $f_n(x)=2n(1/n-x)1_{[0,1/n]}(x).$ Then $int_0^1 |f_n(x)|dx=1$ but $f_n(0)=2n.$
answered Nov 29 at 21:52
John_Wick
1,356111
answered Nov 29 at 21:52
John_Wick
1,356111
answered Nov 29 at 21:52
John_Wick
1,356111
answered Nov 29 at 21:52
John_Wick
1,356111
1,356111
add a comment |
add a comment |
For every $a>0$, the function
$$f_a(x)=frac{2 a e^{-a^2 x^2}}{sqrt{pi} textrm{erf}(a)}$$ with the error function $textrm{erf}(a)=frac{2}{pi}int_0^a e^{-t^2}dt$ is in the set ${|f(0)|: int_{0}^{1}|f(t)|dt=1}$ and evaluates to $f_a(0)=frac{2 a }{sqrt{pi} textrm{erf}(a)}$. Because $textrm{erf}(a) rightarrow 1$ as $arightarrow infty$, we obtain $f_a(0)$ arbitrarily large as we increase $a$. As a consequence, $f_a(0)$ and thus your set are unbounded.
For every $a>0$, the function $f_a(x)$ is in $C[0,1]$ and even infinitely differentiable on $[0,1]$.
answered Nov 29 at 21:53
M1183
943
add a comment |
For every $a>0$, the function
$$f_a(x)=frac{2 a e^{-a^2 x^2}}{sqrt{pi} textrm{erf}(a)}$$ with the error function $textrm{erf}(a)=frac{2}{pi}int_0^a e^{-t^2}dt$ is in the set ${|f(0)|: int_{0}^{1}|f(t)|dt=1}$ and evaluates to $f_a(0)=frac{2 a }{sqrt{pi} textrm{erf}(a)}$. Because $textrm{erf}(a) rightarrow 1$ as $arightarrow infty$, we obtain $f_a(0)$ arbitrarily large as we increase $a$. As a consequence, $f_a(0)$ and thus your set are unbounded.
For every $a>0$, the function $f_a(x)$ is in $C[0,1]$ and even infinitely differentiable on $[0,1]$.
answered Nov 29 at 21:53
M1183
943
add a comment |
For every $a>0$, the function
$$f_a(x)=frac{2 a e^{-a^2 x^2}}{sqrt{pi} textrm{erf}(a)}$$ with the error function $textrm{erf}(a)=frac{2}{pi}int_0^a e^{-t^2}dt$ is in the set ${|f(0)|: int_{0}^{1}|f(t)|dt=1}$ and evaluates to $f_a(0)=frac{2 a }{sqrt{pi} textrm{erf}(a)}$. Because $textrm{erf}(a) rightarrow 1$ as $arightarrow infty$, we obtain $f_a(0)$ arbitrarily large as we increase $a$. As a consequence, $f_a(0)$ and thus your set are unbounded.
For every $a>0$, the function $f_a(x)$ is in $C[0,1]$ and even infinitely differentiable on $[0,1]$.
answered Nov 29 at 21:53
M1183
943
For every $a>0$, the function
$$f_a(x)=frac{2 a e^{-a^2 x^2}}{sqrt{pi} textrm{erf}(a)}$$ with the error function $textrm{erf}(a)=frac{2}{pi}int_0^a e^{-t^2}dt$ is in the set ${|f(0)|: int_{0}^{1}|f(t)|dt=1}$ and evaluates to $f_a(0)=frac{2 a }{sqrt{pi} textrm{erf}(a)}$. Because $textrm{erf}(a) rightarrow 1$ as $arightarrow infty$, we obtain $f_a(0)$ arbitrarily large as we increase $a$. As a consequence, $f_a(0)$ and thus your set are unbounded.
For every $a>0$, the function $f_a(x)$ is in $C[0,1]$ and even infinitely differentiable on $[0,1]$.
answered Nov 29 at 21:53
M1183
943
answered Nov 29 at 21:53
M1183
943
answered Nov 29 at 21:53
M1183
943
answered Nov 29 at 21:53
M1183
943
943
add a comment |
add a comment |
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As written, this makes little sense. What is the precise definition of the set? There's no condition placed on anything. Is it a set of functions?
– Alex R.
Nov 29 at 21:32
This is not a set... you need to assume something about $int_0^1 | f(t)|dt$. I mean it should be a true/false statement...
– Yanko
Nov 29 at 21:32
Is the RHS trying to say that $f$ is absolutely integrable on $[0,1]$? If so, you have the right idea, consider the function $1/sqrt{x}$ and its integral on $[0,1]$
– rubikscube09
Nov 29 at 21:35
Sorry for the confusion. Just edited the description to make it clearer.
– bbw
Nov 29 at 21:37
1
Is it really a problem? What about a constant coefficient?
– Jean-Claude Arbaut
Nov 29 at 21:51