Question about L2 which is not L-infinity

[anselcoffee]

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!

 

[anselcoffee]

bump!

 

[D H]

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?

 

[anselcoffee]

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?

 

[Dick]

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.

 

[D H]

[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 L_p, 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.

 

[anselcoffee]

@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...

****

 

[Dick]

Think polar coordinates if you are working on the disk.

 

[anselcoffee]

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 L_p, 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.