Functional Analysis 13 | Bounded Operators

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This is my video series about Functional Analysis where we start with metric spaces, talk about operators and spectral theory, and end with the famous Spectral Theorem. I hope that it will help everyone who wants to learn about it.

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00:00 Introduction
01:40 Definition - bounded operator
04:35 Proposition - continuous equivalent to bounded

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I hope that this helps students, pupils and others. Have fun!

(This explanation fits to lectures for students in their first and second year of study: Mathematics for physicists, Mathematics for the natural science, Mathematics for engineers and so on)

[xwyl]

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).

[prateekpatel]

Sorry i didnt really get why is operator not called a function ?

[madeleinvanstraaten]

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.

[harrisonbennett]

This series serves as a fantastic comprehensive revision of undergraduate material, excellent!

[PunmasterSTP]

Bounded operators? More like "unBelievable options"...for learning different branches of mathematics! Thanks again so much for making series on so many different topics.

[aqqmamwas]

Great work! Clear and step-by-step explanation of the topic!

[qiaohuizhou]

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?

[gustavocardenas]

Awesome explanation, as always. Thank you very much for the hard work you're doing!

[wlee_mat]

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?

[scollyer.tuition]

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?

[trondsaue]

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 ?

[arcstur]

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?

[darckox]

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?

[mathemavie]

SALUT SVP Y-IL LA TRDUCTION EN FRANCAIS.

[kiwanoish]

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?

[sugarkaylee]

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.

[scollyer.tuition]

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)

[ahmedamr]

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||?

[tmjz]

Simply brilliant. Thank you for all of your help, your videos are always fantastic.