Question about L2 which is not L-infinity

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In summary, the author is looking for a function that is unbounded on or within the unit sphere in R3 but the square of which is integrable over the unit sphere. However, because the domain is limited to a unit sphere, they are worried about some integrals of the function becoming finite.
  • #1
anselcoffee
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Homework Statement


Construct a function u of the space H'(B), where B is a unit sphere in R^3, which does not belong to L∞(B).


Homework Equations



(relevant facts)
all L2 are hilbert, therefore the problem reduces for us finding an L2 function which is not L-infinity.


The Attempt at a Solution


u=((1/(x+1))^(1/c))-1where c>2.

please note that this is adjusted because our domain is limited to a unit sphere. the unadjusted equation looks like this:

u=(1/x)^(1/c) where c>2.

we don't even know if this is correct. spent 2 days thinking about the problem. any help in finding this function would be much appreciated. thank you!
 
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  • #2
bump!
 
  • #3
Don't just guess! You want to find some function that, over a given domain, does not satisfy the definition of membership in [itex]L_{\infty}[/itex] space but does satisfy the definition of membership in [itex]L_2[/itex] space. As a hint, [itex]L_2[/itex] space comprises the set of square-integrable functions over that domain. What, in a nutshell, characterizes [itex]L_{\infty}[/itex] space?
 
  • #4
D H said:
Don't just guess! You want to find some function that, over a given domain, does not satisfy the definition of membership in [itex]L_{\infty}[/itex] space but does satisfy the definition of membership in [itex]L_2[/itex] space. As a hint, [itex]L_2[/itex] space comprises the set of square-integrable functions over that domain. What, in a nutshell, characterizes [itex]L_{\infty}[/itex] space?

[itex]L_{\infty}[/itex] = inf{C>=0: |f(x)| <=C for almost every x}
f is an element of [itex]L_{\infty}[/itex]=||f||p as p approaches infinity is finite

from here, we derived the fact that the function u that we stated is of a p-series with p<1 for L2, therefore, the integration is finite. therefore is an element of L2.

on the other hand, the function u we stated above is of p-series with p>1 for ||f||p as p approaches infinity. therefore its integration is infinite, therefore is not an element of [itex]L_{\infty}[/itex].

i just want second opinions. i just think our analysis of our function u has a lot of loopholes, given the domain constraints (unit sphere).

can you suggest a function that satisfies the condition?
 
  • #5
Forget the details. You can think of L_infinity as just 'bounded'. So you want a square integrable function on the sphere that is unbounded? Should be easy enough. I don't get your examples though. What's x? If you are working on a ball, it might be easier to use spherical coordinates.
 
  • #6
anselcoffee said:
[itex]L_{\infty}[/itex] = inf{C>=0: |f(x)| <=C for almost every x}
In short, a function is in L-infinity if it is bounded throughout the domain except possibly for a set of measure zero. So, you are looking for a function that is unbounded on or within the unit sphere in R3 but the square of which is integrable over the unit sphere.

Some questions:
  • Why are you so bothered by the domain constraints?
    This is good old R3, so you are simply looking at a volume integral over the unit sphere.
  • What do you mean by x in "u=((1/(x+1))^(1/c))-1 where c>2"?
    • If you mean [itex]x\in {\mathcal R}^3, ||x||<=1[/itex], then what does 1/(x+1) mean?
    • If you mean x, as in x,y,z, you should say so.
  • Why do you need to subtract 1?
    For a finite domain, if f(x) is in Lp, then so is g(x)=f(x)+c.

Hint: That the domain is the unit sphere in R3 suggests a spherically symmetric function, thereby reducing the problem to R1.
 
  • #7
@dick

i was just thinking in terms of x,y to make things easier to imagine. i restricted my domain as a unit circle in the x,y plane.

an unbounded f that is square integrable on a unit sphere...

****

thinking thinking thinking...

f(x,y)=exp(1/xy)?

?? it is unbounded as x,y approaches 0.

still thinking...

****
 
  • #8
Think polar coordinates if you are working on the disk.
 
  • #9
D H said:
In short, a function is in L-infinity if it is bounded throughout the domain except possibly for a set of measure zero. So, you are looking for a function that is unbounded on or within the unit sphere in R3 but the square of which is integrable over the unit sphere.

Some questions:
  • Why are you so bothered by the domain constraints?
    This is good old R3, so you are simply looking at a volume integral over the unit sphere.

  • What do you mean by x in "u=((1/(x+1))^(1/c))-1 where c>2"?
    • If you mean [itex]x\in {\mathcal R}^3, ||x||<=1[/itex], then what does 1/(x+1) mean?
    • If you mean x, as in x,y,z, you should say so.

  • Why do you need to subtract 1?
    For a finite domain, if f(x) is in Lp, then so is g(x)=f(x)+c.

Hint: That the domain is the unit sphere in R3 suggests a spherically symmetric function, thereby reducing the problem to R1.

i'm just so concerned because some integrals of functions become finite because of the domain restrictions...

ok. i'll try thinking of a spherically symmetric func, to make things simpler.
 

1. What is the difference between L2 and L-infinity?

L2 and L-infinity are both types of norms used in mathematics and statistics to measure the size or magnitude of a vector or a function. The main difference between them is how they define the "size" of a vector or function. L2 norm is also known as the Euclidean norm and it measures the distance between the origin and the point represented by the vector or function. L-infinity norm, on the other hand, measures the maximum absolute value of the vector or function. In simpler terms, L2 norm focuses on the overall magnitude of the vector while L-infinity norm focuses on the biggest component of the vector.

2. When should I use L2 norm and when should I use L-infinity norm?

The choice between L2 and L-infinity norm depends on the problem you are trying to solve. L2 norm is commonly used when you want to minimize the overall error or distance between two points. It is also used in problems where the data is normally distributed. L-infinity norm is useful when you want to focus on the maximum error or the worst-case scenario. It is commonly used in optimization problems and in situations where the data is not normally distributed.

3. Can L-infinity norm be larger than L2 norm?

Yes, L-infinity norm can be larger than L2 norm. This is because L-infinity norm focuses on the maximum absolute value of the vector or function, while L2 norm takes into account the overall magnitude of the vector or function. In some cases, the biggest component of the vector or function can have a larger absolute value than the overall magnitude of the vector, leading to a larger L-infinity norm.

4. What is the relationship between L2 and L-infinity norms?

L2 and L-infinity norms are two different ways of measuring the size or magnitude of a vector or function. However, they are related in the sense that L-infinity norm can be seen as the limit of L2 norm as the number of dimensions or components of the vector or function increases. In other words, as the dimensionality increases, the maximum absolute value of the vector or function becomes more important and the L-infinity norm becomes a better approximation of the L2 norm.

5. How are L2 and L-infinity norms used in machine learning?

L2 and L-infinity norms are commonly used in machine learning as regularization techniques to prevent overfitting and improve the generalization performance of models. Regularization involves adding a penalty term to the loss function of a model, and L2 and L-infinity norms are often used as the penalty terms. L2 norm is more commonly used as it encourages the model to have smaller weights, while L-infinity norm can be used to enforce sparsity in the model by penalizing large weights. These regularization techniques help prevent overfitting and improve the performance of machine learning models on unseen data.

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