Thanks so much for making it very easy for me to study bounded operators etc! Your explanations are really simple and I can understand in a short time.
In the proof of (b) -> (c), the delta-epsilon definition of continuity almost coincides with the definition of boundedness, i.e. the largest amplification of T is just sup(epsilon/delta).
I was thinking the same, but how can you be sure that sup(epsilon/delta) is not infinity? Therefore, I think, a fixed choice for epsilon has to be made. Or I am wrong?
Bounded operators? More like "unBelievable options"...for learning different branches of mathematics! Thanks again so much for making series on so many different topics.
Given your comment at 10:45, that unbounded linear operators can only occur with an infinite dimensional vector space, I guess that we can conclude that *all* linear operators over finite dimensional spaces (which I suppose is equivalent to saying all matrices) are continuous?
I very much enjoy this series ! One question, though: To elaborate on the statement "T is continuous at x=0" you are proposing to use an epsilon-delta definition. If we generalize to some arbitrary point x0, the definition would read that for every epsilon > 0 there is a delta>0 such that ||x-x0|| < delta implies ||Tx-Tx0||_Y < epsilon. For x0 = 0 this reduces to ||x||_X < delta implies ||Tx||_Y < epsilon. However, you choose apparently epsilon = 1. Is there a particular reasoning behind this ?
@@brightsideofmaths he is trying to say you didnt motivate the reason why you chose 1. asking him what number he would choose doesn't solve the reasoning why you chose 1
@@abublahinocuckbloho4539 There is just no deeper reason for choosing 1. He can choose another number. As long as it is small enough, the proof will also work.
Hi, thx for the fascinating video! I have a questions: Is the proof of claim shown at around 7:22 actually a one-direction proof for the claim in the previous video,i.e. the sequential continuity implies continuity in a metric space?. I know the implication only holds at x=0, but I think if we want to show that any sequentially continuous function is continuous we can prove by contraposition as what was done in this video?
Hi, I always appreciate your videos! I just noticed one thing at 08:08 during the proof of (b)->(c). You tried to prove the contraposition ¬(c)-> ¬(b), but ended up proving (c)->(b). Or did I miss something?
@@brightsideofmaths I hope that when you introduce new concept each video, at least you can tell briefly wy we should care about one concept. This will motivate student alot. Anw, i turn on notification all of your video since i love functional analysis a lot :)
@@HuyNguyen-fp7oz Yeah, I get your point and I really considered a long motivation. However, I decided to get out the facts first and that I can use a whole other video for applications and so on. Sorry if this was not the right call here.
I just have a question, at minute 8:05 you were proving by contraposition, but you use the hypothesis that if the norm of x is less than delta, then the norm of Tx is less than 1. Wouldn't we have to use the hypothesis that |||Tx_n||_Y>1?
Hello my friend, and thank you so much for this series of videos! I have a question: all operators need to be linear + bounded? Or there are operators that are not linear, for example?
Of course, there are also linear maps that are not bounded and there are maps that are not linear. Depending on the context you could call them "operators" but I won't do that here. If I say "operator", I mean a *linear* map.
Thank you for really good videos (I sometimes wish they were even longer and more involved perhaps). Question: At 10:00 you say that we can show that ||T|| is a norm in the usual sense, but what exactly is the usual sense? Everywhere I've looked (e.g. Wikipedia), a norm is defined to be a function to the non-negative real numbers, not to the extended real numbers, i.e. only bounded operators would have norms that are norms in the usual sense. Do we mess anything else up by allowing norms to evaluate to \infty?
@@brightsideofmaths Yes, I understand, my question was whether we let norms take on the value \infty in general? I've never seen that in any definition of norm. So, all norm properties might be satisfied, except the one saying that a norm is a real function.
Am I right in thinking that the operator norm is actually a norm on the vector space of linear operators between two spaces X and Y? If so, is there any reason that you didn't mention it? It would seem to be a significant fact (to me, at least)
@@brightsideofmaths Yes, well, it made me think a bit, for sure. Is the fact that it is a norm important in itself, or is it just an interesting but trivial fact? By the way, I note that at the end, you mention something about "a norm in the usual sense" or similar - I guess that you mean a norm in the sense that I described? I'm afraid that I hadn't actually listened to the whole video before posting my comment.
@@scollyer.tuition Yeah, you are right. I mention it at the end. And I really think that one should try to show the norm properties. Of course, it is a useful fact because it means that you have a new interesting normed space.
An immediate corollary of the proposition is the existence of discontinuous/unbounded linear map. For those interested, Wikipedia provides a cool example: en.wikipedia.org/wiki/Discontinuous_linear_map#A_concrete_example
Thank you so much for making this series of videos! I just have a small question at 5:54. Does the fact that T is linear operator have to mean that Tx_n will go to zero? It seems to me that according to your definition of the linear operator, it could also go to something else.
Thank you so much for your reply! Just a follow-up question: why is it 0 that Tx_n converges to? It is not natural to me because when I think about the example like a linear function f(x)=a*x+b. When x goes to zero, f(x) can go to b, which is not necessarily zero. Maybe I missed something, but I struggle to understand why does that have to be zero.
@@sugarkaylee123 Thanks for the details in your question. A linear function is not a linear map, in general. It is an unfortunate thing that we call a*x+b *linear* instead of *affine* *linear* . A linear map always sends 0 to 0.
Excuse me for my many questions, you're videos are so good! :) Q1: Isn't the completeness of X used in the last proof an additional assumption we did not have in the proposition? Q2: Is the continuity of operator T defined by the continuity of ||T||?
This series serves as a fantastic comprehensive revision of undergraduate material, excellent!
Thanks a lot :)
Yeah I don't think I would consider this or any functional analysis as undergrad material.
@InfiniteQuest86 this certainly was for my undergraduate lectures
Thanks so much for making it very easy for me to study bounded operators etc! Your explanations are really simple and I can understand in a short time.
Glad I could help!
Great work! Clear and step-by-step explanation of the topic!
In the proof of (b) -> (c), the delta-epsilon definition of continuity almost coincides with the definition of boundedness, i.e. the largest amplification of T is just sup(epsilon/delta).
I was thinking the same, but how can you be sure that sup(epsilon/delta) is not infinity? Therefore, I think, a fixed choice for epsilon has to be made. Or I am wrong?
Because T is linear the fraction 2/delta is independent of the choice of epsilon. Is this right? @The Bright Side of Mathematics
Bounded operators? More like "unBelievable options"...for learning different branches of mathematics! Thanks again so much for making series on so many different topics.
Awesome explanation, as always. Thank you very much for the hard work you're doing!
Thank you very much :)
Sorry i didnt really get why is operator not called a function ?
It's called an operator because the map often acts on spaces of functions. It would be confusing to call it also a function then.
Given your comment at 10:45, that unbounded linear operators can only occur with an infinite dimensional vector space, I guess that we can conclude that *all* linear operators over finite dimensional spaces (which I suppose is equivalent to saying all matrices) are continuous?
Exactly :)
@@brightsideofmaths You seem to be on youtube duty 24 hours a day. When do you get time to make the videos :-)
is at 4:10 no 10:45
neatly explained, thanks
I very much enjoy this series ! One question, though: To elaborate on the statement "T is continuous at x=0" you are proposing to use an epsilon-delta definition. If we generalize to some arbitrary point x0, the definition would read that for every epsilon > 0 there is a delta>0 such that ||x-x0|| < delta implies ||Tx-Tx0||_Y < epsilon. For x0 = 0 this reduces to ||x||_X < delta implies ||Tx||_Y < epsilon. However, you choose apparently epsilon = 1. Is there a particular reasoning behind this ?
Thanks! For the proof, we can set epsilon to 1. This does the work :)
@@brightsideofmaths Ok ! So this is apparently a somewhat arbitrary choice.
@@trondsaue7860 Which number would you choose? :)
@@brightsideofmaths he is trying to say you didnt motivate the reason why you chose 1. asking him what number he would choose doesn't solve the reasoning why you chose 1
@@abublahinocuckbloho4539 There is just no deeper reason for choosing 1. He can choose another number. As long as it is small enough, the proof will also work.
Simply brilliant. Thank you for all of your help, your videos are always fantastic.
Hi, thx for the fascinating video! I have a questions:
Is the proof of claim shown at around 7:22 actually a one-direction proof for the claim in the previous video,i.e. the sequential continuity implies continuity in a metric space?. I know the implication only holds at x=0, but I think if we want to show that any sequentially continuous function is continuous we can prove by contraposition as what was done in this video?
Be careful, in this video we only talk about linear maps.
Hi, I always appreciate your videos!
I just noticed one thing at 08:08 during the proof of (b)->(c). You tried to prove the contraposition ¬(c)-> ¬(b), but ended up proving (c)->(b).
Or did I miss something?
We proved (b) -> (c) and we used a proof by contraposition with the two star claims :)
@@brightsideofmaths But the inequality (at 08:19) holds due to ||Tx||
@@wlee_mat Yes
But, why we need bounded operator?
A lot of videos covering this are coming :)
@@brightsideofmaths
I hope that when you introduce new concept each video, at least you can tell briefly wy we should care about one concept. This will motivate student alot. Anw, i turn on notification all of your video since i love functional analysis a lot :)
@@HuyNguyen-fp7oz Yeah, I get your point and I really considered a long motivation. However, I decided to get out the facts first and that I can use a whole other video for applications and so on. Sorry if this was not the right call here.
Thank you very much
You are welcome
I just have a question, at minute 8:05 you were proving by contraposition, but you use the hypothesis that if the norm of x is less than delta, then the norm of Tx is less than 1. Wouldn't we have to use the hypothesis that |||Tx_n||_Y>1?
The proof of green star implies red star is already done. In the next step, we use that the property of red star holds :)
Hello my friend, and thank you so much for this series of videos! I have a question: all operators need to be linear + bounded? Or there are operators that are not linear, for example?
Of course, there are also linear maps that are not bounded and there are maps that are not linear. Depending on the context you could call them "operators" but I won't do that here. If I say "operator", I mean a *linear* map.
@@brightsideofmaths Ohh, understood. Thank you!! :-)
Thank you for really good videos (I sometimes wish they were even longer and more involved perhaps). Question: At 10:00 you say that we can show that ||T|| is a norm in the usual sense, but what exactly is the usual sense? Everywhere I've looked (e.g. Wikipedia), a norm is defined to be a function to the non-negative real numbers, not to the extended real numbers, i.e. only bounded operators would have norms that are norms in the usual sense. Do we mess anything else up by allowing norms to evaluate to \infty?
You have to check the norm properties :)
@@brightsideofmaths Yes, I understand, my question was whether we let norms take on the value \infty in general? I've never seen that in any definition of norm. So, all norm properties might be satisfied, except the one saying that a norm is a real function.
@@kiwanoish It's a usual norm on the space of bounded operators :)
Am I right in thinking that the operator norm is actually a norm on the vector space of linear operators between two spaces X and Y?
If so, is there any reason that you didn't mention it? It would seem to be a significant fact (to me, at least)
Yes, it is a norm! I didn't mention it because either you already know it or you think about it, which is always good :)
@@brightsideofmaths Yes, well, it made me think a bit, for sure. Is the fact that it is a norm important in itself, or is it just an interesting but trivial fact?
By the way, I note that at the end, you mention something about "a norm in the usual sense" or similar - I guess that you mean a norm in the sense that I described? I'm afraid that I hadn't actually listened to the whole video before posting my comment.
@@scollyer.tuition Yeah, you are right. I mention it at the end. And I really think that one should try to show the norm properties. Of course, it is a useful fact because it means that you have a new interesting normed space.
An immediate corollary of the proposition is the existence of discontinuous/unbounded linear map. For those interested, Wikipedia provides a cool example: en.wikipedia.org/wiki/Discontinuous_linear_map#A_concrete_example
Thank you so much for making this series of videos! I just have a small question at 5:54. Does the fact that T is linear operator have to mean that Tx_n will go to zero? It seems to me that according to your definition of the linear operator, it could also go to something else.
Thanks! Yeah, just linearity is not enough for the convergence. We use (b) here, which means the continuity at 0.
Thank you so much for your reply! Just a follow-up question: why is it 0 that Tx_n converges to? It is not natural to me because when I think about the example like a linear function f(x)=a*x+b. When x goes to zero, f(x) can go to b, which is not necessarily zero. Maybe I missed something, but I struggle to understand why does that have to be zero.
@@sugarkaylee123 Thanks for the details in your question. A linear function is not a linear map, in general. It is an unfortunate thing that we call a*x+b *linear* instead of *affine* *linear* . A linear map always sends 0 to 0.
Very clear, thank you!
SALUT SVP Y-IL LA TRDUCTION EN FRANCAIS.
Excuse me for my many questions, you're videos are so good! :)
Q1: Isn't the completeness of X used in the last proof an additional assumption we did not have in the proposition?
Q2: Is the continuity of operator T defined by the continuity of ||T||?
Thanks! For Q1: where did we use the completeness?
For Q2: the continuity of T is defined with respect to the corresponding norms in X and Y.
Q1: You're right, we didn't! I got confused :D
Thanks a lot@@brightsideofmaths