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Define L((0,0)) = {f : (0,0) + R : f is bounded and for every c € (0,0), lim f(z) exists in R} ITc Give an example of & function f € L((0,0)) such t...

Question

Define L((0,0)) = {f : (0,0) + R : f is bounded and for every c € (0,0), lim f(z) exists in R} ITc Give an example of & function f € L((0,0)) such that f is not continuous. Of course you have to prove all your assertions: We equip L((0,0)) with the norm |lfllo sup Ifk) cC(0,0) b. Prove that L((0,0) is a Banach space_ Hint: Use Theorem 7.9 and argue in a similar way as in Theorem 10.4_ You may use that lim f (z) exists ctc if and only if (f(Tn))nz1 is Cauchy for every (Tn)nz1 with Tn €

Define L((0,0)) = {f : (0,0) + R : f is bounded and for every c € (0,0), lim f(z) exists in R} ITc Give an example of & function f € L((0,0)) such that f is not continuous. Of course you have to prove all your assertions: We equip L((0,0)) with the norm |lfllo sup Ifk) cC(0,0) b. Prove that L((0,0) is a Banach space_ Hint: Use Theorem 7.9 and argue in a similar way as in Theorem 10.4_ You may use that lim f (z) exists ctc if and only if (f(Tn))nz1 is Cauchy for every (Tn)nz1 with Tn € (0,c) and limn-o Tn



Answers

Let $(s_{n})$ be a sequence such that \[








|s_{n+1}-s_{n}| < 2^{-n} \qquad \text{for all $n \in \mathbb{N}$.}






\]






Prove $(s_{n})$ is a Cauchy sequence and hence a convergent sequence.

In the problem we have This integration 0 to a fx. Just relax. General S X dx. Now this is equal to what. So this is equal to let us consider this as you And this as which is. So this is of the form islet that is inverse logarithmic algebraic, programmatic and exponential. So let us assume you as the first function and we s the second Frandsen according to this. So we have this is effects immigration do double ds X dx minus integration. D upon dx off fx integration. General does X. Dicks hold X. So further this can be read in us Effects judas X minus integration. This is F. S. X. Into judas X. Dicks. Further we have to solve this spot. So it becomes F L. S X. Division G D S X dx minus integration. Day upon the excessive every day. Six integration Due to six DX holy X. So this is equal to every six and two jx minus integration if they were last X into gx dicks. So this is the different integration of the spot. So these every sex into JD six. The ecstatic was this one. So further we have the digit of overall integration. So this is overall integration will become now first of all we neglected so we neglect if true and G row as these are equals to zero. So what all integration become integration and you go to a fx G doubled us X dx at equals two F f. A judas A minus. So this is the stamp F or fe into Judith a minus. This is if does uh, g of a minus integration zero to A. If nevertheless X G of x dx. So this is the overall integration. Hence it is the answer to the problem.

Suppose you want to find functions F and G. Such that F. S container was at zero and the value of F zero is zero and F G is not continuous at zero. Since F G of X is the same as F of X time strong backs then to make this not continue us, we want to make function G of X That is not continuous at zero since this will affect the product. So how do we make it this container was we just recall that continuity of a function at a point. Let's say X equals C. This is only possible if it Within the three conditions. So the first condition will be for the function G of C to exist. And then the limit as X approaches, see of this function G fx also exist. And then lastly, for this limit as X approaches C of G of X to equal the value of G etc. So if any one of the following do not hold, then it makes it this continuous. So so if you let F of X equal to X squared. That's just container was at zero. Also the value of F at zero here is zero. Then we can make a G of X. A rational expression in which the exponents in the denominator is larger than the exponents of F. So we can say one over X. Cube. So since it didn't say anything about G, then we can form it like this because when you multiply say F G of X, which is the same as F of X G of X. We will then get x squared times one over X cube, which is equal to one over X, which is not continue was at zero. Therefore our functions are x squared and the other one is one over X cube. Get also used one over X to the fourth. As long as it's exponent in the denominator, Fergie is greater than the exponent of F of X. So that when we multiply them, we will still have an X in the denominator, which will Make It. This container was at zero.

In this question here were given. Uh, I am equal to the B N minus. B a minus one. And we're interested in the submission that I am from one option infinity. So notice that they will explain this one when any could you want him to be one minus B zero, then plus B two minus, uh, be one plus the B three months, B two plus the big four minus P five plus up to the PM minus P m minus one. And so, um, and we see them, we can console the B one window. Be one here, be to win the B two B 300 b three. So we're missing the beat three years. So sorry. So let me in. Center doesn't before, uh, be 1234 minus b three here. And then we again this one will be canceled with this one and so on. And here we see that we will cancel this and so on on. We see we have left with only, uh if we consider only up to in the past with some s and for now, then we just stop up to here and then we see that this s and we ego Thio actually end here on me and then we see that this sn it will equal to the We have left with the minus p zero then plus with a p n. And we see that this is here as an we turned the limit here Angus t infinity And then he coaches limit on the industry Infinite. They on the minus B plus B n And we say this one we get echoed you This one will be p zero. So it will be the minus p zero and this leap plus the limit on the PM and goes to infinity. I doesn't implies that the I the submission on the I m converge even only if this limit here also. Uh, I already mitt because you some constant else model that infinity

In this problem were given that a sequence essen is such that The magnitude of difference between the end plus one term and the entire term is always less than do you raise to the power of negative and so we can consider the maximum difference rather the magnitude of the maximum difference. Mhm. Being equal to s to the Brother, 2 to the power negative end. Now a cautious sequence is defined as a sequence in which the subject subsequent terms become arbitrarily close to each other. So If a seven is a sequence than the magnitude of a plus and And a seven plus 1 minus a seven, mhm is equal to zero as N approaches infinity. So they become arbitrarily close to each other. Now to prove that this sequence is a cautious sequence. We can dig the limit as N tends to infinity on both sides. Yeah. Mhm Okay, okay, Okay. Mhm. What people saw limit uh and tends to infinity of due to the negative end and we can solve for the limit. What the fuck? Mhm. Okay. And it's actually a pretty easy limit to solve. So that's the limit as intends to infinity of two to the end. Now, as N approaches infinity, due to the end also approaches infinity, so and is in the in the exponent and as and increases due to the end blows up to infinity and that means that one over to the end. Yeah, Ghost zero. So our limit evaluates 20. Yeah. So the limit as n tends to infinity As 7-plus 1 minus S seven Rather their magnitude equal zero and therefore as intends to infinity the subsequent terms right, subsequent terms Of the sequence, S seven become arbitrarily close, become are bit rarely close, and therefore the given sequence is a cause she sequence. And since cautious sequences always can converge so cause she sequences in words follows that the sequence S N. Don't forget that also. Yeah. Okay. And bridges and there is our solution to this broad.


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