I like your videos but I noticed that you very often say "by the dominated convergence theorem..." I bet it is trivial most of the time, but it would be nice if you could show the dominating function from time to time.
@@Bayerwaldler I totally agree, I am for sure taking advantage a little bit... I'll make a deal, next time I use this theorem I will show the dominating function.
I'm sure there are some interesting examples and then also some trivial examples (like sine) that would both be worth the time to mention the dominating function. I say you probably want to prove the theorem itself once, so any questions can be met with a video link. Same with Fubini and anything you want to do with exchanging the order of two sums or integrals. There is NOTHING worse as a math student than hearing "this works and I won't show you why" Edit to be a little less coy: I am a big fan, and I would love a video about what happens to the bounds when sums and integrals are exchanged, and why it is okay to do.
Hello, Analysis! I can’t believe that we meet again since 2012. Hope you are doing well. I remember what you did to me but now there’s Dr. P you won’t be able to do the same again. Ok, bye.
Thanks for the video man, I just finished my ODE lecture and you starting with a bunch of the basics for PDEs right now is just a blessing. You are a great teacher.
12:50 A more rigorous explanation of why we can replace n goes to infinity with h goes to 0 is that f(h) converges to y as h goes to 0 is equivalent to f (h_n) goes to y for any sequence (h_n) where h_n goes to 0 as n goes to infinity.
My favorite sequence of functions where you can't interchange the limit and integral is f_n(x) = 1 for x in [n, n+1) and 0 otherwise. I think I heard it called the "boxcar" sequence. It's like you're chasing after a runaway boxcar of area 1 to take its integral but it keeps rolling away.
Thank you for sharing this video and the explanation is amazing. This somewhat reminds me of Line integral and how under another function, we find the integral. Is there a relationship between the two?
Thank you very much! Brings back memories of one of my most fun maths class. By the way, that's the first time I see you use computer whiteboard, what is your setup for that? Onenote with mouse and keyboard :p ?
Subsequent question: you seem very chill about the hypothesis verification of the DCT. I don't think that's carelessness, which means that the DCT covers most cases in a nwarly obvious way. Is there a set of functions that are very useful in order to bound function limits? Could you apply the DCT to the pathological case of point wise function you give as an example N*Ind(0,1/N), or give more context? I feel like I am missing a puzzle piece
Hmm..... lets say we have a function: g(x) = (x^5) + (x^4) + (x^2) and a function f(x) = (x^5) + (x^3) + x then for x going to + infinity the difference g(x)-f(x) = x^5 - x^5 + x^4 - x^3 + x^2 - x = 0 + x^3 (x-1) + x (x-1) = (x-1)*x* (x^2 -1) for very large positive x..... Perhaps the ratio would be more informative? f(x)/g(x) = [x * (x^4 + x^2 + 1)] / [ (x^2) * (x^3 + x^2 + 1)] ?
Hey, Dr. P. Nice video, but one nitpick. I would avoid reusing the same letters/identifiers in different and contradictory ways (as you do here with f, g, and c). This would avoid introducing unnecessary confusion (particularly for those who are learning this for the first time). There are a lot of other letters to choose from instead! Cheers! :-D
But Dirac delta is not a function, it is a so called distribution. I know: Physicists used to (or maybe they still do?) call it the "Delta function" but mathematically speaking it is not a function.
The functions are all 0 at 0. If they were n on [0,1/n) instead of (0,1/n), they would converge to the Dirac delta. But then they wouldn't converge pointwise, since f[n](0) goes to ∞.
@@ashuthoshbharadwaj6703 Even in the case Pierre mentioned, the integral of the limit is still 0 as long as you view the limit as a function which can take infinity as a value too (measure theory allows for suchs functions). The idea is that althought f is infinity at 0, the "width of the corresponding rectangle" (or rigorously the measure of the set which contains only 0) is 0, thus the region that is under f is just a vertical line at 0, which ofc has area 0. Is like integrating 0 over all the reals, here the width is infinite but the height is 0 and as we all know the integral is 0. Now if you go distributions'-land, that's another story...
So in Analysis 2 we studied passing the limit under the integral sign with uniform convergence. Does the Dominated Convergence Theorem apply to more general functions?
When you are talking about integral, I assume you mean in the sense of Riemann. If that is the case, then you need to add the additional assumption that the pointwise limiting function f is Riemann-integrable as well.
9:40 Why don’t we just say, lim l f_n(x) l < infinity? Rather than using g(x). What would be a counter example when lim f_n is finite, lim Integral f_n =/ integral lim f ?
Question: Why do we assume the f_n converges to 0 point wise, as by the definition of the indicator function as n -> inf f_n (x) = 0 (expect zero), for zero it seems (intuitively) [n -> inf ] n -> f_n(0) = inf. This seems to be like a delta function (For which the integral would be 1?), it will be really helpful if you can explain / point towards an explanation of what i am missing. Thanks a lot for this awesome videos.
@@di-dah-ditcharlie6511 even in the case you mentioned the integral of the limit function is still 0, because it might be infinity at a point, but that only gives us a vertical line under the graph of f, which still has area 0. Is like integrating 0 from 0 to infinity, the region under the graph here is the x-axis, which ofc has area 0. That being said, that's how riemann and lebesgue integrals work. There are other interpretations which may give integral = 1 as you expected (check out distributions for more details on that)
Is the absolute value around f_N(x) necessary when we are comparing it to some other function g(x) for each N within the statement of the Dominated Convergence Theorem?
@@drpeyam Thank you. I may have answered my own question initially after posting this comment with what I wrote and appreciate the example to help clarify and reinforce my original intuition. I look forward to more videos :).
DCT theorem is only for Lebesgue integrals?Is it Lebesgue theorem? And if I want to apply Lebesgue ( Dominated convergence Theorem)sentence in combination with MVT, I need to make a measurable space was interval and functions on this interval continuous, is MVT Lagrange ?
Technically the domain could be any measure space, which is one of the many reasons Lebesgue integral is so powerful (another reason is the DCT itself). Also it is enough to suppose that the domain of f is a subset of the domain of g (which is kinda obvious if you draw the graphs).
If the limit function f is bounded and we are considering functions on a set with finite measure, then uniform convergence implies dominated convergence for n sufficiently large, since then we have f(x)-e
@@Happy_Abe It doesn't though. Forget about what the sequence looks like graphically and focus on the definition of pointwise convergence; the limit of the sequence [f_1(0), f_2(0), f_3(0), ...] is 0. Therefore f(0) = 0 and f is not the Dirac delta distribution.
I am psyched you made this video! I use this theorem in my videos all the time but haven't really wanted to prove it!
I like your videos but I noticed that you very often say "by the dominated convergence theorem..." I bet it is trivial most of the time, but it would be nice if you could show the dominating function from time to time.
@@Bayerwaldler I totally agree, I am for sure taking advantage a little bit... I'll make a deal, next time I use this theorem I will show the dominating function.
I'm sure there are some interesting examples and then also some trivial examples (like sine) that would both be worth the time to mention the dominating function. I say you probably want to prove the theorem itself once, so any questions can be met with a video link. Same with Fubini and anything you want to do with exchanging the order of two sums or integrals. There is NOTHING worse as a math student than hearing "this works and I won't show you why"
Edit to be a little less coy: I am a big fan, and I would love a video about what happens to the bounds when sums and integrals are exchanged, and why it is okay to do.
You should both prove it :( yours and Dr. peyams's are my favorite youtube channels
its still not proved. please do!
Hello, Analysis!
I can’t believe that we meet again since 2012. Hope you are doing well. I remember what you did to me but now there’s Dr. P you won’t be able to do the same again.
Ok, bye.
Hahahaha
after studying 10years of pure math, I'm 100% sure DCT is my fav theorem! the assumptions are so mild that you can't pass on loving it!
Thanks for the video man, I just finished my ODE lecture and you starting with a bunch of the basics for PDEs right now is just a blessing. You are a great teacher.
Using the Dirac Delta Distribution as a countexample for a property of a function is a bit evil! :-P
More videos of this kind please, Dr. PEyam.
8:40
I chuckled at the Dominican Central Time part, and I'm half Dominican! Great analysis video.
12:50 A more rigorous explanation of why we can replace n goes to infinity with h goes to 0 is that f(h) converges to y as h goes to 0 is equivalent to f (h_n) goes to y for any sequence (h_n) where h_n goes to 0 as n goes to infinity.
Definitely the most unknowingly used theorem I’ve come across.
Smooth and brilliant
this whole time, finally a worthy explanation
This comes in really handy, writing my german Analysis II exam on friday.
Viel Glück!!!
okay this makes sense now i understand whats going on with the PDE course
Thank you so much.
A fantastic video! Thanks Peyam!
Yass! the video I was looking for
😄
This is way better than my video’s coverage of the DCT! Not that I’m complaining. It’s really, really good.
finally someone made a video about this
Grande Dr peyam. Lo más grande de todo el mundo.
Good. Thank you very much.
greate video with very clear explanation, thank you very much
I have to see this video a couple of times.
My favorite sequence of functions where you can't interchange the limit and integral is f_n(x) = 1 for x in [n, n+1) and 0 otherwise. I think I heard it called the "boxcar" sequence. It's like you're chasing after a runaway boxcar of area 1 to take its integral but it keeps rolling away.
I like that!!!
well done
Thank you for sharing this video and the explanation is amazing. This somewhat reminds me of Line integral and how under another function, we find the integral. Is there a relationship between the two?
This sounds like it would be useful for proving properties of the discrete cosine transform, but it would be really confusing namewise.
Good explanation
The fundamental theorem of analysis!
Thank you very much! Brings back memories of one of my most fun maths class.
By the way, that's the first time I see you use computer whiteboard, what is your setup for that? Onenote with mouse and keyboard :p ?
Subsequent question: you seem very chill about the hypothesis verification of the DCT. I don't think that's carelessness, which means that the DCT covers most cases in a nwarly obvious way. Is there a set of functions that are very useful in order to bound function limits? Could you apply the DCT to the pathological case of point wise function you give as an example N*Ind(0,1/N), or give more context? I feel like I am missing a puzzle piece
I’m using the microsoft whiteboard + zoom
And here we cannot bound n 1(0,1/n) because otherwise we could pass in the limit!
Hmm..... lets say we have a function: g(x) = (x^5) + (x^4) + (x^2) and a function f(x) = (x^5) + (x^3) + x then for x going to + infinity the difference g(x)-f(x) =
x^5 - x^5 + x^4 - x^3 + x^2 - x = 0 + x^3 (x-1) + x (x-1) = (x-1)*x* (x^2 -1) for very large positive x.....
Perhaps the ratio would be more informative?
f(x)/g(x) = [x * (x^4 + x^2 + 1)] / [ (x^2) * (x^3 + x^2 + 1)] ?
Hey, Dr. P. Nice video, but one nitpick. I would avoid reusing the same letters/identifiers in different and contradictory ways (as you do here with f, g, and c). This would avoid introducing unnecessary confusion (particularly for those who are learning this for the first time). There are a lot of other letters to choose from instead! Cheers! :-D
Wouldnt the example function converge to a dirac delta?? It had a constant area of 1 for all N
But Dirac delta is not a function, it is a so called distribution. I know: Physicists used to (or maybe they still do?) call it the "Delta function" but mathematically speaking it is not a function.
@@Bayerwaldler Yeah that makes sense I guess, but I'm still not convinced with calling it a zero everywhere, it has to be infinite at x = 0
The functions are all 0 at 0. If they were n on [0,1/n) instead of (0,1/n), they would converge to the Dirac delta. But then they wouldn't converge pointwise, since f[n](0) goes to ∞.
@@pierreabbat6157 Ahhh thats it! Thank you so much for clearing that up!
@@ashuthoshbharadwaj6703 Even in the case Pierre mentioned, the integral of the limit is still 0 as long as you view the limit as a function which can take infinity as a value too (measure theory allows for suchs functions). The idea is that althought f is infinity at 0, the "width of the corresponding rectangle" (or rigorously the measure of the set which contains only 0) is 0, thus the region that is under f is just a vertical line at 0, which ofc has area 0. Is like integrating 0 over all the reals, here the width is infinite but the height is 0 and as we all know the integral is 0. Now if you go distributions'-land, that's another story...
finally!
Is it a complete proof?
👌 thx
So in Analysis 2 we studied passing the limit under the integral sign with uniform convergence. Does the Dominated Convergence Theorem apply to more general functions?
This is more general. If fn converges uniformly to f, then it is bounded by an integrable function g (I think)
When you are talking about integral, I assume you mean in the sense of Riemann. If that is the case, then you need to add the additional assumption that the pointwise limiting function f is Riemann-integrable as well.
No, Lebesgue
@@drpeyam Ah, okay! In that case, good content!
9:40
Why don’t we just say, lim l f_n(x) l < infinity? Rather than using g(x).
What would be a counter example when lim f_n is finite, lim Integral f_n =/ integral lim f ?
Well you could just define g(x) as lim |fn(x)| so it’s equivalent, but it’s important that g be integrable
Dr Peyam
Thank you professor!
Question: Why do we assume the f_n converges to 0 point wise, as by the definition of the indicator function as n -> inf f_n (x) = 0 (expect zero), for zero it seems (intuitively) [n -> inf ] n -> f_n(0) = inf. This seems to be like a delta function (For which the integral would be 1?), it will be really helpful if you can explain / point towards an explanation of what i am missing. Thanks a lot for this awesome videos.
It’s because we’re using the open internal (0,1/n). If we used the closed interval [0,1/n] then I’d agree with you
@@drpeyam Thanks, I get it now.
@@di-dah-ditcharlie6511 even in the case you mentioned the integral of the limit function is still 0, because it might be infinity at a point, but that only gives us a vertical line under the graph of f, which still has area 0. Is like integrating 0 from 0 to infinity, the region under the graph here is the x-axis, which ofc has area 0. That being said, that's how riemann and lebesgue integrals work. There are other interpretations which may give integral = 1 as you expected (check out distributions for more details on that)
Is the absolute value around f_N(x) necessary when we are comparing it to some other function g(x) for each N within the statement of the Dominated Convergence Theorem?
Yes because fn could be negative. Think for instance -n 1(0,1/n). Then it is bounded above by 0 but the DCT doesn’t hold there
@@drpeyam Thank you. I may have answered my own question initially after posting this comment with what I wrote and appreciate the example to help clarify and reinforce my original intuition. I look forward to more videos :).
DCT theorem is only for Lebesgue integrals?Is it Lebesgue theorem?
And if I want to apply Lebesgue ( Dominated convergence Theorem)sentence in combination with MVT, I need to make a measurable space was interval and functions on this interval continuous, is MVT Lagrange ?
Yes, Lebesgue integrals. And yes, you’d need it to be differentiable etc
@@drpeyam Yes, thank you.
So since you didn't mention the domain in the theorem, I assume that means that the domain of g is the same as f and is a subset of R?
Yeah
Technically the domain could be any measure space, which is one of the many reasons Lebesgue integral is so powerful (another reason is the DCT itself). Also it is enough to suppose that the domain of f is a subset of the domain of g (which is kinda obvious if you draw the graphs).
Is this theorem eqivavalet to uniform convergence of f_n, and if not does it imply uniform convergence
If the limit function f is bounded and we are considering functions on a set with finite measure, then uniform convergence implies dominated convergence for n sufficiently large, since then we have f(x)-e
Aleksandra Bozovic thanks that was helpful
Why isn't the limit of the function a delta function?
The delta "function" isn't a function. It's something called a distribution.
@@martinepstein9826 I still have my question then, it should converge to that distribution
@@Happy_Abe It doesn't though. Forget about what the sequence looks like graphically and focus on the definition of pointwise convergence; the limit of the sequence [f_1(0), f_2(0), f_3(0), ...] is 0. Therefore f(0) = 0 and f is not the Dirac delta distribution.
@@martinepstein9826 I was focusing on the graphical element yeah, thanks.
16:04 Mi español interior ha despertado 😆
8:36 Why is there a theorem about Dominican Central Time?
8:39 Oh nvmd
LOL
So beautiful 💕👍👍
noice