||f|| >= 0 because the squared function will be non-negative, so the sum (integral) will also be non-negative and so will the square root.
The ||f|| will be 0 only if each of the values inside the integral are 0 which means f(x)^2 takes the value 0 at all points from 0 to 1, which essentially means f(x) is 0. Question: But, what if f(x) has values other than 0, outside the interval [0,1]. Won't the norm still be 0, but the function is not the "zero function"?
||af|| = a||f|| This is true because multiplying f with a, squaring and integrating it would have a^2 as a constant and then taking the root of that would just be a that can be pulled out of the expression leaving us with a||f||.
Squaring both sides gives us an extra term on the L.H.S. that is >= 0
keenan
Question: But, what if f(x) has values other than 0, outside the interval [0,1]. Won't the norm still be 0, but the function is not the "zero function"?
When we talk about a function being defined over [0,1], it does not have values outside this interval. It may help to take a step back and think about it more abstractly: in general, a "function" associates to each point in a set X a single value in a set Y. For instance, consider the sets X = { dog, lizard, cat, turtle } and Y = { green, black, brown }. Then we might have a function f(dog) = brown, f(lizard) = green, f(cat) = black, and f(turtle) = green. The function f is not defined for any values (in this case, animals!) other than those in X.
Likewise, when we say f is a function with domain [0,1], we really mean it! In other words, we really mean that it cannot be evaluated on any points other than [0,1].
keenan
There is still a subtlety to (2) that is not being addressed. Anyone have any ideas?
tso
Can (2) be interpreted with the help of inner product?
Essentially, ||f||^2 = <f,f>, which means the norm is the square root of the inner product. Now we know that a function should perfectly line up with itself. Thus, the norm can only be 0 if f(x) = 0
haojun
Does the function have to be a continuous one? If not, (2) may not be strictly satisfied. For example, if f(0)=15462,f(1)=15662,f(x)=0 for 0<x<1, I think the norm is still 0.
haojun
If the number of points where f(x)!=0 is enumerable, while at other points f(x)=0, the norm is still 0, I think.
Of course, f(x) must not equal to infinity at any point.
If f(x)=15662 for all rational numbers between 0 and 1, and f(x)=0 for irrational ones, the norm might still be 0.
michell3
Is there a reason L^2 norm is called that?
johnryan
^ This is what I was thinking. The slide doesn't specify whether the square of the function needs to have a well-defined Riemann integral, or Lebesgue integral, etc. I actually think the number of points where the function is nonzero may be uncountable, and still the norm could be zero, as long as f(x)=0 "almost everywhere". For example, the indicator function of the Cantor set (https://en.wikipedia.org/wiki/Cantor_set). Hopefully this doesn't appear on a test!
johnryan
@michell3 The L is for Lebesgue, who was interested in measuring things. The 2 is for the fact that we're SQUARING the integrand, and taking the SQUARE root of the integral. In L^3 norm, we CUBE the integrand, and take the CUBE root of the integral.
More here: https://en.wikipedia.org/wiki/Norm_(mathematics)#p-norm
keenan
Does the function have to be a continuous one?
Right—this is the key question. Nowhere did we say the function has to be continuous. So, it can be nonzero on a set of measure zero and still have a zero integral. (This stuff will also NOT be on the test! :-)) In this case, one might call $||\cdot||$ a "seminorm" rather than a norm. Another example is
$$ ||f|| := f(0), $$
which satisfies all of the properties of a norm except the condition that only the zero function has zero norm. (More generally, any norm restricted to a subset of the domain will behave this way.)
This discussion may sound a bit pedantic/abstract, but actually this "functional analytic" viewpoint is something people adopt quite a lot in modern image and surface processing, and becomes especially important when you start talking about sparse phenomena like the edges in an image (which we discussed briefly in class).
rftucker
So now I'm a little confused about the formal definition of a norm. The formal definition slide says that if the norm of vector v is equal to 0, then bijectively we can say that v is the zero vector. But if f(x) does not have to be continuous necessarily, then does that mean the definition of a norm has to be different depending on whether or not we are defining it for vector functions vs scalar functions?
xpfbob
I'm little confused about the meaning of L^2 norm. For example, we have two different functions on the interval [0,1], it's possible that they have different shapes but have the same L^2 norm. So we can say they have the same L^2 norm, it's same big, but actually it can behave very differently on [0,1]. So what does it mean?
johnryan
@rftucker The statement that the norm of a vector being zero is equivalent to the vector being the zero vector is still true, due to the following technicality. In $L^2$, vectors which differ on a set of measure zero are considered to be the same vector. Therefore, the function which is zero on the irrationals but 1 on the rationals (which has $L^2$ norm = 0) is EQUAL to the function f(x)=0 - they are the same vector. Using this part of the definition, the norm properties are satisfied, and indeed there is still only one zero vector. More on this here http://mathworld.wolfram.com/L2-Space.html.
I'm not sure what you mean by vector functions vs. scalar functions, do you mean different depending on whether we are defining it for vectors like <2,1,4> vs. for functions like f(x)=2x-1? If that is what you mean, then the formal definition slide applies to all cases.
@xpfbob Is it a problem that two vectors can have the same norm? Think about the vectors <1, 0> and <0,1>. They are much different vectors, but have the same norm, right? The same can happen with function in $L^2$.
The only case where two different vectors having the same norm is a problem is if that norm is 0. Why is this a problem? Well, the straight answer is that the definition of a norm requires that only one vector have norm zero. The reason for this a priori arbitrary rule is that, if multiple different vectors could have norm zero, then you'd have bad math problems (imagine if the real numbers had two different numbers as zero!)
keenan
@xpfbob The situation you describe is no different from what happens in $\mathbb{R}^n$: consider two vectors (3,4,5) and (-4,3,-5). They are not the same vector, but have exactly the same norm. Same deal with functions: there may be many many functions with the same norm, even though the functions themselves look quite different. This outcome is no big surprise: you've basically summarized an entire function with a single number—so no wonder this number does not tell you everything about the function!
||f|| >= 0 because the squared function will be non-negative, so the sum (integral) will also be non-negative and so will the square root.
The ||f|| will be 0 only if each of the values inside the integral are 0 which means f(x)^2 takes the value 0 at all points from 0 to 1, which essentially means f(x) is 0. Question: But, what if f(x) has values other than 0, outside the interval [0,1]. Won't the norm still be 0, but the function is not the "zero function"?
||af|| = a||f|| This is true because multiplying f with a, squaring and integrating it would have a^2 as a constant and then taking the root of that would just be a that can be pulled out of the expression leaving us with a||f||.
Squaring both sides gives us an extra term on the L.H.S. that is >= 0
When we talk about a function being defined over [0,1], it does not have values outside this interval. It may help to take a step back and think about it more abstractly: in general, a "function" associates to each point in a set X a single value in a set Y. For instance, consider the sets X = { dog, lizard, cat, turtle } and Y = { green, black, brown }. Then we might have a function f(dog) = brown, f(lizard) = green, f(cat) = black, and f(turtle) = green. The function f is not defined for any values (in this case, animals!) other than those in X.
Likewise, when we say f is a function with domain [0,1], we really mean it! In other words, we really mean that it cannot be evaluated on any points other than [0,1].
There is still a subtlety to (2) that is not being addressed. Anyone have any ideas?
Can (2) be interpreted with the help of inner product?
Essentially, ||f||^2 = <f,f>, which means the norm is the square root of the inner product. Now we know that a function should perfectly line up with itself. Thus, the norm can only be 0 if f(x) = 0
Does the function have to be a continuous one? If not, (2) may not be strictly satisfied. For example, if f(0)=15462,f(1)=15662,f(x)=0 for 0<x<1, I think the norm is still 0.
If the number of points where f(x)!=0 is enumerable, while at other points f(x)=0, the norm is still 0, I think.
Of course, f(x) must not equal to infinity at any point.
If f(x)=15662 for all rational numbers between 0 and 1, and f(x)=0 for irrational ones, the norm might still be 0.
Is there a reason L^2 norm is called that?
^ This is what I was thinking. The slide doesn't specify whether the square of the function needs to have a well-defined Riemann integral, or Lebesgue integral, etc. I actually think the number of points where the function is nonzero may be uncountable, and still the norm could be zero, as long as f(x)=0 "almost everywhere". For example, the indicator function of the Cantor set (https://en.wikipedia.org/wiki/Cantor_set). Hopefully this doesn't appear on a test!
@michell3 The L is for Lebesgue, who was interested in measuring things. The 2 is for the fact that we're SQUARING the integrand, and taking the SQUARE root of the integral. In L^3 norm, we CUBE the integrand, and take the CUBE root of the integral.
More here: https://en.wikipedia.org/wiki/Norm_(mathematics)#p-norm
Right—this is the key question. Nowhere did we say the function has to be continuous. So, it can be nonzero on a set of measure zero and still have a zero integral. (This stuff will also NOT be on the test! :-)) In this case, one might call $||\cdot||$ a "seminorm" rather than a norm. Another example is
$$ ||f|| := f(0), $$
which satisfies all of the properties of a norm except the condition that only the zero function has zero norm. (More generally, any norm restricted to a subset of the domain will behave this way.)
This discussion may sound a bit pedantic/abstract, but actually this "functional analytic" viewpoint is something people adopt quite a lot in modern image and surface processing, and becomes especially important when you start talking about sparse phenomena like the edges in an image (which we discussed briefly in class).
So now I'm a little confused about the formal definition of a norm. The formal definition slide says that if the norm of vector v is equal to 0, then bijectively we can say that v is the zero vector. But if f(x) does not have to be continuous necessarily, then does that mean the definition of a norm has to be different depending on whether or not we are defining it for vector functions vs scalar functions?
I'm little confused about the meaning of L^2 norm. For example, we have two different functions on the interval [0,1], it's possible that they have different shapes but have the same L^2 norm. So we can say they have the same L^2 norm, it's same big, but actually it can behave very differently on [0,1]. So what does it mean?
@rftucker The statement that the norm of a vector being zero is equivalent to the vector being the zero vector is still true, due to the following technicality. In $L^2$, vectors which differ on a set of measure zero are considered to be the same vector. Therefore, the function which is zero on the irrationals but 1 on the rationals (which has $L^2$ norm = 0) is EQUAL to the function f(x)=0 - they are the same vector. Using this part of the definition, the norm properties are satisfied, and indeed there is still only one zero vector. More on this here http://mathworld.wolfram.com/L2-Space.html. I'm not sure what you mean by vector functions vs. scalar functions, do you mean different depending on whether we are defining it for vectors like <2,1,4> vs. for functions like f(x)=2x-1? If that is what you mean, then the formal definition slide applies to all cases.
@xpfbob Is it a problem that two vectors can have the same norm? Think about the vectors <1, 0> and <0,1>. They are much different vectors, but have the same norm, right? The same can happen with function in $L^2$. The only case where two different vectors having the same norm is a problem is if that norm is 0. Why is this a problem? Well, the straight answer is that the definition of a norm requires that only one vector have norm zero. The reason for this a priori arbitrary rule is that, if multiple different vectors could have norm zero, then you'd have bad math problems (imagine if the real numbers had two different numbers as zero!)
@xpfbob The situation you describe is no different from what happens in $\mathbb{R}^n$: consider two vectors (3,4,5) and (-4,3,-5). They are not the same vector, but have exactly the same norm. Same deal with functions: there may be many many functions with the same norm, even though the functions themselves look quite different. This outcome is no big surprise: you've basically summarized an entire function with a single number—so no wonder this number does not tell you everything about the function!