It's funny, I encountered so many proofs along these lines in my point set topology class and I always found them so dry and hence difficult to follow (or rather, difficult to stay awake long enough to follow them). However, the visuals in this video make it so clear what we're referring to, and the gory details are just there for if you get lost and need some extra structure. Having the visuals as the main proof and the details on the side is delightful, and I really enjoyed this proof. The whole thing seemed borderline obvious, which just goes to show you've done a great job presenting it if your audience is able to guess the conclusion ahead of time.
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I love the smooth jazz on the background. Makes the learning experience much better, enjoyable and calmer.
Thanks very much! Quick back story, I started my YT channel to animate this exact proof! I drew it all out in one note trying to make sense of what was going on (it was written out in words in the textbook!) and thought hmm that's actually worth sharing!
Thx. i joined one of the Commentators ( maxwel wilbert) saying : a* is not the good approximation( bc. it is still far away from the point f. if you want to get a good approx. to f, you consider Sub-Space A like a Ballon. As it gets bigger & bigger changes the geometry ( perhaps going from an Appel into a Pear inorder to get as close as possible to f.) At the same time Space B , shrinks toward f , the soulution changes to " minimax " of John v. Newman. Happy to hear more on that from you & the fans.
@@Your_choise Well a series that is actually locally taylor expandable (ie every point has a neighborhood where the function is given by a taylor series) must be smooth (all derivatives of any order exist). Is the function infinitely often differentiable? No: Not taylor expandable. Yes: You can thus form the taylor series by virtue of those derivatives and check the radius of convergence. Is the radius positive? No: Oops, taylor doesn't work. Yes: Does the taylor series actually converge to the original function on that point? No: Oops, no Taylor for you. Yes: Great! Taylor worked. For example consider the function f:R->R given by e^(-1/x^2) for x!=0 and 0 for x=0. This function is infinitely often differentiable everywhere (even 0) and the taylor series at 0 exists and converges everywhere. But since all derivatives at x=0 are 0 the taylor series is identicially the 0-function so does NOT agree with f anywhere except x=0!
I promise if you keep pushing with this and find ways to take your brilliant explanations and compound them into some final, cool application you will most certainly blow up. You are so lucid in how you articulate things it's hard to put it into words how things just click when you explain them.
@@Eta_Carinae__ how "near" or "far" two functions from each other depends on the norm? yeah i just never thought you can even define a "distance" between something that's not a point in some space
@@geoffrygifari3377 Norn is just a real value function with three properties, and a metric in this case is also a function with three properties. if you have a norm, there is a metric induces from this norm in the sense we can measure distance between two element of the set (now function is an element or "Point"of the set of functions). Its actually not a point in normal sense.
Enjoyed the video. As someone who works in ML, its sad to know there exists a best approximation out there but you are stuck with what your model learned.
Also, it would be nice to note that, in this particular case (with the L² norm and A being a vector space), an explicit formula for the best approximation can be derived. It turns out a* is the orthogonal projection of f onto A. It can be computed by projecting f onto an orthogonal basis of A. If the approximation is meant to cover the interval [0, 1], the basis could be the first three shifted Legendre polynomials: 1, 2x−1, and 6x²−6x+1. See for instance the articles “Orthogonal polynomials” and “Legendre polynomials” on Wikipedia.
Numerically, A being compact is not enough, it also needs to be *located*, i.e. in addition to upper bounds you need to be able to find a lower approximation of the distance from a point to the set in order to find a good approximation of d*. Then this is closely related to Bishop's lemma in constructive analysis
The set of at most degree 2 polynomials is not compact, but do form a closed convex non-empty subset in a Hilbert space of L^2(X) where X is a bounded subset of the some Euclidean space. Therefore these minimizing elements exist.
I think in practical terms, since linear regression minimises the square of distance from data points to approximating function, then a good measure of how good an approximation is would be to integrate the square of difference between a function and its approximation over the finite range we want (a finite Taylor series will shoot out to infinity eventually).
This is really cool. I'll like the abstract nature of metric space. But how does this help find 'the best approximation' to the function y=e^-x? How could finding a* help in this case?
Thanks! I think the proof that the best approximation a* exists doesn't actually help you find it . In practice, it will depend on the method, and the method will have its own norm (measure of distance). so for example least-squares approximation might have a different a* to minimax approximation (which minimises the maximum distance between the 2 functions). I guess the proof is here to show finding the best approximation is worth doing. In a sense its the supporting theory behind the algorithms! Its useful to compare to eg. differential equations: its great if we can prove a solution exists even if it doesn't help us solve it because at least we're not wasting our time with something which will turn out to have no solutions!
@@DrWillWood Thanks for your answer :) I find it fascinating that an objective subject like mathematics still seems quite subjective about what algorithm to use to perform the 'best' fit. You could use a least squares, you could also use a nonlinear least squares, or polynomial fit etc etc. The criteria for 'best' seems to change depending on the algorithm.
@@matthewjames7513 That subjectivity is fully captured within your distance function. The reason to use metric spaces here is it means that the theorem works for any definition of "best"
If you know your model is in the form of a*e^-(b*x)+c, it is easy to find the coefficients 'a','b', and 'c' by using an "optimizing" or "minimizing" routine such as BFGS, or Levenberg-Marquardt. There are many 'optimizing' methods. Most try to minimize the sum of squared errors over a range. If you have only two coefficients to find, 'a' and 'b', then you can think of this as 'a' the north south location and 'b' as the east west location of your first guess at 'a' and 'b' . The sum of squared errors between your data and your model a*e^-(b*x) is the elevation for each point. What you try to do is go down hill to find a lower sum of squared error or elevation. You keep trying to go down hill with each step until you find the elevation or sum of square errors is less the epsilon or you reach a step limit. Eventually you get to the valley or canyon floor if you don't fall into a pit ( local minimum ). If you know there are local minimums then you may need to start 'a' and 'b' at different locations a few times to make sure you find the true minimum. If you have many coefficients then it is as if you are trying to go down hill in multiple dimensions. Learning to minimize errors or optimize is an eye opener.
@@pnachtwey Hey! I've actually done an optimizations course before so my eyes have already been opened haha. I never studied it in the context of a metric space before though. Let's say you wanted to find the parameters a,b, and c in your example. I still find it strange that there is no "best" method to extract those parameters. You could use least squares, you could use LM, you could use fmincon() in MATLAB and they'll all give you similar answers - but how can you know which is the best method?
Nice visuals. This is just for one norm. I wonder if we can do a similar comparison for every possible norm. Create a metric space of norms and see which one gets us the closest to the function.
Thanks you! I think any closed and bounded set in a metric space is compact so I think in this case closed and compact are pretty synonymous :-) edit: oops I got mixed up here! A compact subset of a metric space is closed and bounded not the other way! To be honest, I don't know if the theorem holds if A is closed but not compact but certainly this proof relies on it.
Any convex subset of a Banach (complete) metric space can define a perpendicular projection, and thus distance. One needs closure so that the set contains its limit points.
isn't there a small mistake in your reasoning? at 5:30 you talk about a sequence of points a_i in A such that lim i -> ∞ d(a_i,f)=d*. then you say that this sequence has a limit point a+, which is not necessarily the case if the sequence has multiple cluster points(which will be in A). you first have to choose one of those points and then take that for a+. for example you have the compact set [0,1]U[3,4] in R(euclidian metric). you want to approximate 2 and have the sequence a_n= 1-(1/2)^n for n even , and a_n=3+(1/2)^n for n odd. this sequence doesn`t converge, but lim i -> ∞ d(a_i,2)=1. now you can choose 1 or 3 (the cluster points) and continue with that point for your a+. edit: and you have to choose a subsequence that actually goes to your chosen a+)
If two points are both with the same distance, they converge to a better aproximation because A is a compact subset of the metric space B. For the case you describe, you would have to aproximate to a subset C of the subset A you are using, therefore, you don't need to aproximate because the point you want to aproximate is already in the subset A. Imagine that the metric space is a rectangle and the subset is any irregular shape "1", logically, for any point outside the irregular shape "1", there exist only one point, which must be in the contour of the shape "1". If there exist a irregular shape "2", subset of the shape "1", there exist many point equally closet to a point in the shape "2" as many of them exist in the inscribed circle of the contour of the shape "2". So one can choose the point in the shape "2" because is subset of the shepa "1".
Maybe I havent done formal math in a while but this proof seems faulty to me. Namely 5:00 is unjustified, you are trying to prove that a* exists but you assume apriori that d* exists. I'm pretty sure the proof is supposed to be as follows: The metric d(x, y) is a continuous function where x, y \in B, therefor d(x, f) is continuous in B, and thus in A. A is compact, thus d restricted to A obtains its minimum (and, maximum) in A.
Yes! This proof is from "approximation theory and methods" by M.J.D. Powell which certainly I would say is a deep delve on the subject! Theres another book "An introduction to the approximation of functions" by T.J. Rivlin which covers similar material but is less thorough.
I wouldn't expect students to come across this material until late undergraduate level! I studied this proof for my masters degree so don't worry if it feels difficult!
I don't understand why at 5:00 when you let d* be the shortest distance from A to f, that you need to prove anything from there. Don't we already have that d*=min{d(a,f): a \in A} by that assumption, and then whatever d* is you can just say there must exist some a in A such that d(a,f)=d*? If there did not exist a in A such that d(a,f) = d*, then every a in A would be some distance other than d* away from f, and thus d* would not be the minimum. But it is the minimum, so that's a contradiction, and thus there has to exist a in A such that d(a,f)=d*. Is this not sufficient to prove existence?
Think of the example with the interval (0,1) and the point 2. First of all, the distance between the interval and the number 2 is 1, since |2-1-e| = |1-e| -> 1 when e -> 0 (e here is epsilon not the number e), and we also know that all points in (0,1) are less than one so the distance is at most 1. Thus the distance is 1. This is however never reached by any element in (0,1). For a proof, let's say for x -x=-1 => x=1. But x
I realize this is just nitpicking but I believe your proof doesn't show that a* is THE optimal solution, only that it is AN optimal solution. Uniqueness is a whole other story
Great video! But I can't be the only one who thinks the topic seems awfully abstract. _"Within a space, there exists a point within a subspace that is closest to some other point outside the subspace."_ It's definitely correct, but since those points could represent literally anything, I don't see how this definition is anything but a truism. There exists a best way to approximate one function with another. Is that not already obvious? What am I missing?
In a space with out a Metric, how would you assess that the approximation is close to the target function in any sense? In the absence of a distance function you can't judge
Not sure but I suspect it might be possible in more general spaces like uniform spaces. They have pseudometrics instead. Since all metrics are pseudometrics, all metric spaces are uniform spaces but not vice versa. Maybe you can find something that works in a uniform space which is not a metric space. In general to answer your type of question (can we work with less mathematical structure? ), I look at the hierarchy of mathematical spaces. If you want to generalize results go to more general spaces and see if the proof still holds with the reduced mathematical structure. From left to right below the spaces have less and less structure and are more and more general. (e.g All Inner Product Spaces are Normed Vector spaces): Inner Product Space, Normed Vector Space, Metric Space, Uniform Space, Cauchy Space, Topological Space, Convergence Space
You mention the L₂ norm but, from what I have read, in approximation theory the goal is usually to minimize the worst case error, i.e. the L∞ norm of a−f.
It's funny, I encountered so many proofs along these lines in my point set topology class and I always found them so dry and hence difficult to follow (or rather, difficult to stay awake long enough to follow them). However, the visuals in this video make it so clear what we're referring to, and the gory details are just there for if you get lost and need some extra structure. Having the visuals as the main proof and the details on the side is delightful, and I really enjoyed this proof. The whole thing seemed borderline obvious, which just goes to show you've done a great job presenting it if your audience is able to guess the conclusion ahead of time.
I love the smooth jazz on the background. Makes the learning experience much better, enjoyable and calmer.
Great video. Thanks! The theoretical value of a subset being compact in a metric space is hugely under appreciated .
dude this is awesome, keep going!!!
thanks a lot! appreciate the support!
That existence proof was great. I needed a refresher for my math course. Thanks for sharing!
Thanks very much! Quick back story, I started my YT channel to animate this exact proof! I drew it all out in one note trying to make sense of what was going on (it was written out in words in the textbook!) and thought hmm that's actually worth sharing!
@@DrWillWood wow that's inspiring
Thx.
i joined one of the Commentators ( maxwel wilbert) saying : a* is not the good approximation( bc. it is still far away from the point f. if you want to get a good approx. to f, you consider Sub-Space A like a Ballon. As it gets bigger & bigger changes the geometry ( perhaps going from an Appel into a Pear inorder to get as close as possible to f.) At the same time Space B , shrinks toward f , the soulution changes to " minimax " of John v. Newman. Happy to hear more on that from you & the fans.
Infinite series are exact equivalents; only truncated series are approximations.
You are right, but his Taylor series was just an example here. That A subset of the function space B can be anything we choose, in general.
There are functions that can't be represented by taylor series.
@@stydras3380 out of curiosity, is there a simple test to test what functions can be modeled by Taylor series
@@Your_choise Well a series that is actually locally taylor expandable (ie every point has a neighborhood where the function is given by a taylor series) must be smooth (all derivatives of any order exist). Is the function infinitely often differentiable? No: Not taylor expandable. Yes: You can thus form the taylor series by virtue of those derivatives and check the radius of convergence. Is the radius positive? No: Oops, taylor doesn't work. Yes: Does the taylor series actually converge to the original function on that point? No: Oops, no Taylor for you. Yes: Great! Taylor worked. For example consider the function f:R->R given by e^(-1/x^2) for x!=0 and 0 for x=0. This function is infinitely often differentiable everywhere (even 0) and the taylor series at 0 exists and converges everywhere. But since all derivatives at x=0 are 0 the taylor series is identicially the 0-function so does NOT agree with f anywhere except x=0!
Function "distance to f" is continuous so by Weierstrass theorem it reaches its infinium on any compact. Q.E.D.
I promise if you keep pushing with this and find ways to take your brilliant explanations and compound them into some final, cool application you will most certainly blow up. You are so lucid in how you articulate things it's hard to put it into words how things just click when you explain them.
I'm used to imagining distance between points but distance between *continuous functions* is completely new
It's kinda arbitrary tbh. To get something meaningful, you have to define a norm, so you could have several different d* depending on the norm you use
@@Eta_Carinae__ how "near" or "far" two functions from each other depends on the norm? yeah i just never thought you can even define a "distance" between something that's not a point in some space
@@geoffrygifari3377 Norn is just a real value function with three properties, and a metric in this case is also a function with three properties. if you have a norm, there is a metric induces from this norm in the sense we can measure distance between two element of the set (now function is an element or "Point"of the set of functions). Its actually not a point in normal sense.
@@thanhminhcao3356 ah math.... where one concept can be generalized to something else in no way seemingly related
Check out 'Topology' by James Munkres.
Enjoyed the video.
As someone who works in ML, its sad to know there exists a best approximation out there but you are stuck with what your model learned.
Loving your videos! Keep up the great work
Also, it would be nice to note that, in this particular case (with the L² norm and A being a vector space), an explicit formula for the best approximation can be derived. It turns out a* is the orthogonal projection of f onto A. It can be computed by projecting f onto an orthogonal basis of A. If the approximation is meant to cover the interval [0, 1], the basis could be the first three shifted Legendre polynomials: 1, 2x−1, and 6x²−6x+1.
See for instance the articles “Orthogonal polynomials” and “Legendre polynomials” on Wikipedia.
Numerically, A being compact is not enough, it also needs to be *located*, i.e. in addition to upper bounds you need to be able to find a lower approximation of the distance from a point to the set in order to find a good approximation of d*. Then this is closely related to Bishop's lemma in constructive analysis
Loving your video Dr. Wood! Great explanations :D
In optical applications there are Zernike coefficents describing optical surfaces based on orthogonal polynoms.
The set of at most degree 2 polynomials is not compact, but do form a closed convex non-empty subset in a Hilbert space of L^2(X) where X is a bounded subset of the some Euclidean space. Therefore these minimizing elements exist.
And due to convexity you have uniqueness too 😁
Thank you very much, Will, your videos are so good and easy to understand!!
I think in practical terms, since linear regression minimises the square of distance from data points to approximating function, then a good measure of how good an approximation is would be to integrate the square of difference between a function and its approximation over the finite range we want (a finite Taylor series will shoot out to infinity eventually).
You made it soo easy to understand 👍
you explained this very clearly and perfectly tnx a lottttt
This is really cool. I'll like the abstract nature of metric space. But how does this help find 'the best approximation' to the function y=e^-x? How could finding a* help in this case?
Thanks! I think the proof that the best approximation a* exists doesn't actually help you find it . In practice, it will depend on the method, and the method will have its own norm (measure of distance). so for example least-squares approximation might have a different a* to minimax approximation (which minimises the maximum distance between the 2 functions). I guess the proof is here to show finding the best approximation is worth doing. In a sense its the supporting theory behind the algorithms! Its useful to compare to eg. differential equations: its great if we can prove a solution exists even if it doesn't help us solve it because at least we're not wasting our time with something which will turn out to have no solutions!
@@DrWillWood Thanks for your answer :) I find it fascinating that an objective subject like mathematics still seems quite subjective about what algorithm to use to perform the 'best' fit. You could use a least squares, you could also use a nonlinear least squares, or polynomial fit etc etc. The criteria for 'best' seems to change depending on the algorithm.
@@matthewjames7513 That subjectivity is fully captured within your distance function. The reason to use metric spaces here is it means that the theorem works for any definition of "best"
If you know your model is in the form of a*e^-(b*x)+c, it is easy to find the coefficients 'a','b', and 'c' by using an "optimizing" or "minimizing" routine such as BFGS, or Levenberg-Marquardt. There are many 'optimizing' methods. Most try to minimize the sum of squared errors over a range.
If you have only two coefficients to find, 'a' and 'b', then you can think of this as 'a' the north south location and 'b' as the east west location of your first guess at 'a' and 'b' . The sum of squared errors between your data and your model a*e^-(b*x) is the elevation for each point. What you try to do is go down hill to find a lower sum of squared error or elevation. You keep trying to go down hill with each step until you find the elevation or sum of square errors is less the epsilon or you reach a step limit. Eventually you get to the valley or canyon floor if you don't fall into a pit ( local minimum ). If you know there are local minimums then you may need to start 'a' and 'b' at different locations a few times to make sure you find the true minimum. If you have many coefficients then it is as if you are trying to go down hill in multiple dimensions. Learning to minimize errors or optimize is an eye opener.
@@pnachtwey Hey! I've actually done an optimizations course before so my eyes have already been opened haha. I never studied it in the context of a metric space before though. Let's say you wanted to find the parameters a,b, and c in your example. I still find it strange that there is no "best" method to extract those parameters. You could use least squares, you could use LM, you could use fmincon() in MATLAB and they'll all give you similar answers - but how can you know which is the best method?
Nice visuals. This is just for one norm. I wonder if we can do a similar comparison for every possible norm. Create a metric space of norms and see which one gets us the closest to the function.
Or you could just say that the function a |---> d(a,f) has attains a minimum because A is compact.
these videos are amazing!
Great video! Quick question: why would this theorem fail if we loosen the restriction on A to simply be closed, rather than compact?
Thanks you! I think any closed and bounded set in a metric space is compact so I think in this case closed and compact are pretty synonymous :-)
edit: oops I got mixed up here! A compact subset of a metric space is closed and bounded not the other way! To be honest, I don't know if the theorem holds if A is closed but not compact but certainly this proof relies on it.
@@leirumf5476 thanks! you're right, got mixed up there! I've put in a correction above :-)
Any convex subset of a Banach (complete) metric space can define a perpendicular projection, and thus distance. One needs closure so that the set contains its limit points.
@@DrWillWood o.o sorry, I just see that. I'mma delete my thingy 🙈
Do all solutions lie on the boundary?
isn't there a small mistake in your reasoning?
at 5:30 you talk about a sequence of points a_i in A such that lim i -> ∞ d(a_i,f)=d*.
then you say that this sequence has a limit point a+, which is not necessarily the case if the sequence has multiple cluster points(which will be in A). you first have to choose one of those points and then take that for a+.
for example you have the compact set [0,1]U[3,4] in R(euclidian metric). you want to approximate 2 and have the sequence a_n= 1-(1/2)^n for n even , and a_n=3+(1/2)^n for n odd. this sequence doesn`t converge, but lim i -> ∞ d(a_i,2)=1. now you can choose 1 or 3 (the cluster points) and continue with that point for your a+.
edit: and you have to choose a subsequence that actually goes to your chosen a+)
What is the significance of two points in A that are both identically close to f while also being the minimum distance?
If two points are both with the same distance, they converge to a better aproximation because A is a compact subset of the metric space B. For the case you describe, you would have to aproximate to a subset C of the subset A you are using, therefore, you don't need to aproximate because the point you want to aproximate is already in the subset A.
Imagine that the metric space is a rectangle and the subset is any irregular shape "1", logically, for any point outside the irregular shape "1", there exist only one point, which must be in the contour of the shape "1". If there exist a irregular shape "2", subset of the shape "1", there exist many point equally closet to a point in the shape "2" as many of them exist in the inscribed circle of the contour of the shape "2". So one can choose the point in the shape "2" because is subset of the shepa "1".
Thank you so much
Maybe I havent done formal math in a while but this proof seems faulty to me. Namely 5:00 is unjustified, you are trying to prove that a* exists but you assume apriori that d* exists. I'm pretty sure the proof is supposed to be as follows: The metric d(x, y) is a continuous function where x, y \in B, therefor d(x, f) is continuous in B, and thus in A. A is compact, thus d restricted to A obtains its minimum (and, maximum) in A.
Exact thing I want to learn. What are some books that delve deep into this sort of math?
Yes! This proof is from "approximation theory and methods" by M.J.D. Powell which certainly I would say is a deep delve on the subject! Theres another book "An introduction to the approximation of functions" by T.J. Rivlin which covers similar material but is less thorough.
very nice intuition video, i like it
Thank you :-)
Senior in high school here, about to take the AP Calculus BC exam (comparable to Calc 2). Should someone like me be expected to understand this?
I wouldn't expect students to come across this material until late undergraduate level! I studied this proof for my masters degree so don't worry if it feels difficult!
I don't understand why at 5:00 when you let d* be the shortest distance from A to f, that you need to prove anything from there. Don't we already have that d*=min{d(a,f): a \in A} by that assumption, and then whatever d* is you can just say there must exist some a in A such that d(a,f)=d*? If there did not exist a in A such that d(a,f) = d*, then every a in A would be some distance other than d* away from f, and thus d* would not be the minimum. But it is the minimum, so that's a contradiction, and thus there has to exist a in A such that d(a,f)=d*. Is this not sufficient to prove existence?
Think of the example with the interval (0,1) and the point 2. First of all, the distance between the interval and the number 2 is 1, since |2-1-e| = |1-e| -> 1 when e -> 0 (e here is epsilon not the number e), and we also know that all points in (0,1) are less than one so the distance is at most 1. Thus the distance is 1. This is however never reached by any element in (0,1). For a proof, let's say for x -x=-1 => x=1. But x
@@MK-13337 That is a simple way to put it, thank you.
I realize this is just nitpicking but I believe your proof doesn't show that a* is THE optimal solution, only that it is AN optimal solution. Uniqueness is a whole other story
Great video! But I can't be the only one who thinks the topic seems awfully abstract. _"Within a space, there exists a point within a subspace that is closest to some other point outside the subspace."_ It's definitely correct, but since those points could represent literally anything, I don't see how this definition is anything but a truism. There exists a best way to approximate one function with another. Is that not already obvious?
What am I missing?
Hello, just wondering.
Does this implies that for non-metric spaces you can’t have a best approximation?
ps. Great video
In a space with out a Metric, how would you assess that the approximation is close to the target function in any sense? In the absence of a distance function you can't judge
Not sure but I suspect it might be possible in more general spaces like uniform spaces. They have pseudometrics instead. Since all metrics are pseudometrics, all metric spaces are uniform spaces but not vice versa. Maybe you can find something that works in a uniform space which is not a metric space.
In general to answer your type of question (can we work with less mathematical structure? ), I look at the hierarchy of mathematical spaces. If you want to generalize results go to more general spaces and see if the proof still holds with the reduced mathematical structure.
From left to right below the spaces have less and less structure and are more and more general. (e.g All Inner Product Spaces are Normed Vector spaces):
Inner Product Space, Normed Vector Space, Metric Space, Uniform Space, Cauchy Space, Topological Space, Convergence Space
you can probably do more or less analogous things with quasimetrics, but I expect a lot of caveats
@@smolboi9659 wouldnt that just be a case of quotienting it out? Id expect the theory would be the same up to quotients
Thankyou.
Closed and bounded isn't the same as compact for a general metric space
thanks for pointing that out, I'll put a clarification about this in the description :-)
@@DrWillWood it's a great video by the way. Wish my real analysis textbook has visuals like that
Closed and bounded are necessary but not sufficient for a general metric space.
what a roundabout and unecessary way of not explaining anything. All notation no content.
You mention the L₂ norm but, from what I have read, in approximation theory the goal is usually to minimize the worst case error, i.e. the L∞ norm of a−f.
Proof looks a bit unsound to me.