Does a bounded norm of a function in L2 mean the function is bounded?Showing that the space $C[0,1]$ with the...
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Does a bounded norm of a function in L2 mean the function is bounded?
Showing that the space $C[0,1]$ with the $L_1$ norm is incompleteHow is $lvert frvert=int^1_0lvert frvert dx$ not a norm (f is a regulated function, or funtion with bounded variation 0)Norm of bounded real functionBounded functions, existence of a functional with a given normCalculating the norm of a bounded linear functionalFunction not satisfying the mean value theoremIs the space of almost everywhere differentiable function with bounded derivative embedded with uniform norm complete?What does bounded mean in this context?Under a finite dimensional norm-induced metric space, is every bounded sequence has a converging subsequence?Usage of mean value theorem ; bounded derivative and open interval
$begingroup$
In $L^2$ space, if the norm of a function $||f||$ is bounded on $[0,1]$ does that mean the function is bounded?
If so, why? Is there a theorem? Or a counter example?
real-analysis
asked Mar 20 at 0:09
FrankFrank
17410
$endgroup$
add a comment |
$begingroup$
In $L^2$ space, if the norm of a function $||f||$ is bounded on $[0,1]$ does that mean the function is bounded?
If so, why? Is there a theorem? Or a counter example?
real-analysis
asked Mar 20 at 0:09
FrankFrank
17410
$endgroup$
$begingroup$
In general, an integral bound on a function cannot give a pointwise bound. It is enough to remark that there are unbounded functions with bounded integral.
$endgroup$
– Beni Bogosel
Mar 20 at 0:17
add a comment |
$begingroup$
In $L^2$ space, if the norm of a function $||f||$ is bounded on $[0,1]$ does that mean the function is bounded?
If so, why? Is there a theorem? Or a counter example?
real-analysis
asked Mar 20 at 0:09
FrankFrank
17410
$endgroup$
In $L^2$ space, if the norm of a function $||f||$ is bounded on $[0,1]$ does that mean the function is bounded?
If so, why? Is there a theorem? Or a counter example?
real-analysis
real-analysis
asked Mar 20 at 0:09
FrankFrank
17410
asked Mar 20 at 0:09
FrankFrank
17410
asked Mar 20 at 0:09
FrankFrank
17410
asked Mar 20 at 0:09
FrankFrank
17410
asked Mar 20 at 0:09
FrankFrank
17410
17410
$begingroup$
In general, an integral bound on a function cannot give a pointwise bound. It is enough to remark that there are unbounded functions with bounded integral.
$endgroup$
– Beni Bogosel
Mar 20 at 0:17
add a comment |
$begingroup$
In general, an integral bound on a function cannot give a pointwise bound. It is enough to remark that there are unbounded functions with bounded integral.
$endgroup$
– Beni Bogosel
Mar 20 at 0:17
$begingroup$
In general, an integral bound on a function cannot give a pointwise bound. It is enough to remark that there are unbounded functions with bounded integral.
$endgroup$
– Beni Bogosel
Mar 20 at 0:17
$begingroup$
In general, an integral bound on a function cannot give a pointwise bound. It is enough to remark that there are unbounded functions with bounded integral.
$endgroup$
– Beni Bogosel
Mar 20 at 0:17
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This is not true. Consider the function
$$
f(x) := begin{cases}
frac{1}{x^{1/4}} & text{if } x in (0,1],\
0 & text{if } x = 0.
end{cases}
$$
Clearly, $f$ is unbounded on $[0,1]$. However,
begin{align*}
leftVert f rightVert_{L^2}^2 = int_0^1 |f|^2,mathrm{d}m = int_0^1 frac{1}{sqrt{x}},mathrm{d}x = 2.
end{align*}
Moreover, as pointed out in the comments, the $L^2$-norm does not give you much information on the pointwise values of a function. Indeed, by the very nature of the Lebesgue integral, changing the values of $f$ on a null set would not affect the value of the integral.
answered Mar 20 at 0:11
rolandcyprolandcyp
2,289422
$endgroup$
$begingroup$
Isn't the norm in $L^2=sqrt{x^2}$? If I am wrong... please tell me why, it would really help me.
$endgroup$
– Frank
Mar 20 at 0:16
$begingroup$
The $L^2([0,1])$-norm of a function $f$ is $left( int_0^1 |f|^2 right)^{1/2}$.
$endgroup$
– rolandcyp
Mar 20 at 0:17
$begingroup$
Can you tell me why? Where can I read that this is the case? When I look up Lp space it tells me the norm $||X||_p$ is $sqrt{x^p}$
$endgroup$
– Frank
Mar 20 at 0:28
$begingroup$
A norm has to return a real number not a function. In fact, since elements of $L^2$ aren't well defined pointwise, your "norm" would not be well defined. Which source are you referring to?
$endgroup$
– rolandcyp
Mar 20 at 0:28
1
$begingroup$
For $1 leq p < infty$, the norm on $L^p(X)$ is simply the map $$ f mapsto left( int_X |f|^p,mathrm{d}mright)^{1/p}.$$ See en.wikipedia.org/wiki/Lp_space#Lp_spaces
$endgroup$
– rolandcyp
Mar 20 at 0:29
|
show 2 more comments
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This is not true. Consider the function
$$
f(x) := begin{cases}
frac{1}{x^{1/4}} & text{if } x in (0,1],\
0 & text{if } x = 0.
end{cases}
$$
Clearly, $f$ is unbounded on $[0,1]$. However,
begin{align*}
leftVert f rightVert_{L^2}^2 = int_0^1 |f|^2,mathrm{d}m = int_0^1 frac{1}{sqrt{x}},mathrm{d}x = 2.
end{align*}
Moreover, as pointed out in the comments, the $L^2$-norm does not give you much information on the pointwise values of a function. Indeed, by the very nature of the Lebesgue integral, changing the values of $f$ on a null set would not affect the value of the integral.
answered Mar 20 at 0:11
rolandcyprolandcyp
2,289422
$endgroup$
$begingroup$
Isn't the norm in $L^2=sqrt{x^2}$? If I am wrong... please tell me why, it would really help me.
$endgroup$
– Frank
Mar 20 at 0:16
$begingroup$
The $L^2([0,1])$-norm of a function $f$ is $left( int_0^1 |f|^2 right)^{1/2}$.
$endgroup$
– rolandcyp
Mar 20 at 0:17
$begingroup$
Can you tell me why? Where can I read that this is the case? When I look up Lp space it tells me the norm $||X||_p$ is $sqrt{x^p}$
$endgroup$
– Frank
Mar 20 at 0:28
$begingroup$
A norm has to return a real number not a function. In fact, since elements of $L^2$ aren't well defined pointwise, your "norm" would not be well defined. Which source are you referring to?
$endgroup$
– rolandcyp
Mar 20 at 0:28
1
$begingroup$
For $1 leq p < infty$, the norm on $L^p(X)$ is simply the map $$ f mapsto left( int_X |f|^p,mathrm{d}mright)^{1/p}.$$ See en.wikipedia.org/wiki/Lp_space#Lp_spaces
$endgroup$
– rolandcyp
Mar 20 at 0:29
|
show 2 more comments
$begingroup$
This is not true. Consider the function
$$
f(x) := begin{cases}
frac{1}{x^{1/4}} & text{if } x in (0,1],\
0 & text{if } x = 0.
end{cases}
$$
Clearly, $f$ is unbounded on $[0,1]$. However,
begin{align*}
leftVert f rightVert_{L^2}^2 = int_0^1 |f|^2,mathrm{d}m = int_0^1 frac{1}{sqrt{x}},mathrm{d}x = 2.
end{align*}
Moreover, as pointed out in the comments, the $L^2$-norm does not give you much information on the pointwise values of a function. Indeed, by the very nature of the Lebesgue integral, changing the values of $f$ on a null set would not affect the value of the integral.
answered Mar 20 at 0:11
rolandcyprolandcyp
2,289422
$endgroup$
$begingroup$
Isn't the norm in $L^2=sqrt{x^2}$? If I am wrong... please tell me why, it would really help me.
$endgroup$
– Frank
Mar 20 at 0:16
$begingroup$
The $L^2([0,1])$-norm of a function $f$ is $left( int_0^1 |f|^2 right)^{1/2}$.
$endgroup$
– rolandcyp
Mar 20 at 0:17
$begingroup$
Can you tell me why? Where can I read that this is the case? When I look up Lp space it tells me the norm $||X||_p$ is $sqrt{x^p}$
$endgroup$
– Frank
Mar 20 at 0:28
$begingroup$
A norm has to return a real number not a function. In fact, since elements of $L^2$ aren't well defined pointwise, your "norm" would not be well defined. Which source are you referring to?
$endgroup$
– rolandcyp
Mar 20 at 0:28
1
$begingroup$
For $1 leq p < infty$, the norm on $L^p(X)$ is simply the map $$ f mapsto left( int_X |f|^p,mathrm{d}mright)^{1/p}.$$ See en.wikipedia.org/wiki/Lp_space#Lp_spaces
$endgroup$
– rolandcyp
Mar 20 at 0:29
|
show 2 more comments
$begingroup$
This is not true. Consider the function
$$
f(x) := begin{cases}
frac{1}{x^{1/4}} & text{if } x in (0,1],\
0 & text{if } x = 0.
end{cases}
$$
Clearly, $f$ is unbounded on $[0,1]$. However,
begin{align*}
leftVert f rightVert_{L^2}^2 = int_0^1 |f|^2,mathrm{d}m = int_0^1 frac{1}{sqrt{x}},mathrm{d}x = 2.
end{align*}
Moreover, as pointed out in the comments, the $L^2$-norm does not give you much information on the pointwise values of a function. Indeed, by the very nature of the Lebesgue integral, changing the values of $f$ on a null set would not affect the value of the integral.
answered Mar 20 at 0:11
rolandcyprolandcyp
2,289422
$endgroup$
This is not true. Consider the function
$$
f(x) := begin{cases}
frac{1}{x^{1/4}} & text{if } x in (0,1],\
0 & text{if } x = 0.
end{cases}
$$
Clearly, $f$ is unbounded on $[0,1]$. However,
begin{align*}
leftVert f rightVert_{L^2}^2 = int_0^1 |f|^2,mathrm{d}m = int_0^1 frac{1}{sqrt{x}},mathrm{d}x = 2.
end{align*}
Moreover, as pointed out in the comments, the $L^2$-norm does not give you much information on the pointwise values of a function. Indeed, by the very nature of the Lebesgue integral, changing the values of $f$ on a null set would not affect the value of the integral.
answered Mar 20 at 0:11
rolandcyprolandcyp
2,289422
edited Mar 20 at 0:19
answered Mar 20 at 0:11
rolandcyprolandcyp
2,289422
answered Mar 20 at 0:11
rolandcyprolandcyp
2,289422
answered Mar 20 at 0:11
rolandcyprolandcyp
2,289422
2,289422
$begingroup$
Isn't the norm in $L^2=sqrt{x^2}$? If I am wrong... please tell me why, it would really help me.
$endgroup$
– Frank
Mar 20 at 0:16
$begingroup$
The $L^2([0,1])$-norm of a function $f$ is $left( int_0^1 |f|^2 right)^{1/2}$.
$endgroup$
– rolandcyp
Mar 20 at 0:17
$begingroup$
Can you tell me why? Where can I read that this is the case? When I look up Lp space it tells me the norm $||X||_p$ is $sqrt{x^p}$
$endgroup$
– Frank
Mar 20 at 0:28
$begingroup$
A norm has to return a real number not a function. In fact, since elements of $L^2$ aren't well defined pointwise, your "norm" would not be well defined. Which source are you referring to?
$endgroup$
– rolandcyp
Mar 20 at 0:28
1
$begingroup$
For $1 leq p < infty$, the norm on $L^p(X)$ is simply the map $$ f mapsto left( int_X |f|^p,mathrm{d}mright)^{1/p}.$$ See en.wikipedia.org/wiki/Lp_space#Lp_spaces
$endgroup$
– rolandcyp
Mar 20 at 0:29
|
show 2 more comments
$begingroup$
Isn't the norm in $L^2=sqrt{x^2}$? If I am wrong... please tell me why, it would really help me.
$endgroup$
– Frank
Mar 20 at 0:16
$begingroup$
The $L^2([0,1])$-norm of a function $f$ is $left( int_0^1 |f|^2 right)^{1/2}$.
$endgroup$
– rolandcyp
Mar 20 at 0:17
$begingroup$
Can you tell me why? Where can I read that this is the case? When I look up Lp space it tells me the norm $||X||_p$ is $sqrt{x^p}$
$endgroup$
– Frank
Mar 20 at 0:28
$begingroup$
A norm has to return a real number not a function. In fact, since elements of $L^2$ aren't well defined pointwise, your "norm" would not be well defined. Which source are you referring to?
$endgroup$
– rolandcyp
Mar 20 at 0:28
1
$begingroup$
For $1 leq p < infty$, the norm on $L^p(X)$ is simply the map $$ f mapsto left( int_X |f|^p,mathrm{d}mright)^{1/p}.$$ See en.wikipedia.org/wiki/Lp_space#Lp_spaces
$endgroup$
– rolandcyp
Mar 20 at 0:29
$begingroup$
Isn't the norm in $L^2=sqrt{x^2}$? If I am wrong... please tell me why, it would really help me.
$endgroup$
– Frank
Mar 20 at 0:16
$begingroup$
Isn't the norm in $L^2=sqrt{x^2}$? If I am wrong... please tell me why, it would really help me.
$endgroup$
– Frank
Mar 20 at 0:16
$begingroup$
The $L^2([0,1])$-norm of a function $f$ is $left( int_0^1 |f|^2 right)^{1/2}$.
$endgroup$
– rolandcyp
Mar 20 at 0:17
$begingroup$
The $L^2([0,1])$-norm of a function $f$ is $left( int_0^1 |f|^2 right)^{1/2}$.
$endgroup$
– rolandcyp
Mar 20 at 0:17
$begingroup$
Can you tell me why? Where can I read that this is the case? When I look up Lp space it tells me the norm $||X||_p$ is $sqrt{x^p}$
$endgroup$
– Frank
Mar 20 at 0:28
$begingroup$
Can you tell me why? Where can I read that this is the case? When I look up Lp space it tells me the norm $||X||_p$ is $sqrt{x^p}$
$endgroup$
– Frank
Mar 20 at 0:28
$begingroup$
A norm has to return a real number not a function. In fact, since elements of $L^2$ aren't well defined pointwise, your "norm" would not be well defined. Which source are you referring to?
$endgroup$
– rolandcyp
Mar 20 at 0:28
$begingroup$
A norm has to return a real number not a function. In fact, since elements of $L^2$ aren't well defined pointwise, your "norm" would not be well defined. Which source are you referring to?
$endgroup$
– rolandcyp
Mar 20 at 0:28
1
1
$begingroup$
For $1 leq p < infty$, the norm on $L^p(X)$ is simply the map $$ f mapsto left( int_X |f|^p,mathrm{d}mright)^{1/p}.$$ See en.wikipedia.org/wiki/Lp_space#Lp_spaces
$endgroup$
– rolandcyp
Mar 20 at 0:29
$begingroup$
For $1 leq p < infty$, the norm on $L^p(X)$ is simply the map $$ f mapsto left( int_X |f|^p,mathrm{d}mright)^{1/p}.$$ See en.wikipedia.org/wiki/Lp_space#Lp_spaces
$endgroup$
– rolandcyp
Mar 20 at 0:29
|
show 2 more comments
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$begingroup$
In general, an integral bound on a function cannot give a pointwise bound. It is enough to remark that there are unbounded functions with bounded integral.
$endgroup$
– Beni Bogosel
Mar 20 at 0:17