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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

1

$begingroup$

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

share|cite|improve this question

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 | 

1

$begingroup$

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

share|cite|improve this question

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

share|cite|improve this question

asked Mar 20 at 0:09

FrankFrank

17410

$endgroup$

share|cite|improve this question

asked Mar 20 at 0:09

FrankFrank

17410

share|cite|improve this question

asked Mar 20 at 0:09

FrankFrank

17410

asked Mar 20 at 0:09

FrankFrank

17410

asked Mar 20 at 0:09

FrankFrank

17410

1 Answer
1

active

oldest

votes

4

$begingroup$

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.

share|cite|improve this answer

edited Mar 20 at 0:19

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

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4

$begingroup$

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.

share|cite|improve this answer

edited Mar 20 at 0:19

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

4

$begingroup$

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.

share|cite|improve this answer

edited Mar 20 at 0:19

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.

share|cite|improve this answer

edited Mar 20 at 0:19

answered Mar 20 at 0:11

rolandcyprolandcyp

2,289422

$endgroup$

share|cite|improve this answer

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