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Problem 1._ We say a function [0, 1] 3 [0. 1] is a line if there are m,b € R such that f(x) =mc+b for all x € [0,1]. Let L([0 , 1] 4 [0, 1]) denote the ...

Question

Problem 1._ We say a function [0, 1] 3 [0. 1] is a line if there are m,b € R such that f(x) =mc+b for all x € [0,1]. Let L([0 , 1] 4 [0, 1]) denote the subspace of C([0,1] ~ [0,1]) with the metrie dx consisting of lines. Show L([O,1] ~ [0,1]) is compact: Hint: First, show that if In(x) mnx + bn, then there are subsequences (mnk JkeN and (bnre )keN that converge in R.

Problem 1._ We say a function [0, 1] 3 [0. 1] is a line if there are m,b € R such that f(x) =mc+b for all x € [0,1]. Let L([0 , 1] 4 [0, 1]) denote the subspace of C([0,1] ~ [0,1]) with the metrie dx consisting of lines. Show L([O,1] ~ [0,1]) is compact: Hint: First, show that if In(x) mnx + bn, then there are subsequences (mnk JkeN and (bnre )keN that converge in R.



Answers

Let f : [a,b] to R be an integrable function. Define G : [a,b] to R by G(x) = integral x to b f(t)dt. Show that G is comtinuous on [a,b]. Further, if f is continuous at c belonging [a,b], then G is diffrentiable at c and G'(x) = -f(c).

Yeah. So in this problem we need to sure that any bounded decreasing sequence of real numbers converges. So we are given about a decreasing sequence. We named that A. N. And we need to show that A. N. Can gorgeous to what limit? We don't care about this. Okay, so let's start with the proof, send the sequence in is bounded in particular. It does bounded below that is the set A. And such that and comes from natural numbers is a bounded subset five. Have we known that are has completeness property also known as and you'll be property that is least upper bound property. And this is equal into G. L. B. Property also known as greatest lower bound property, basically this means, but if any subset of R is bounded below, then it must have an infamy. Um So in fema of the scent and coming from and coming from national numbers is some real number. Let's call this L. So our claim is like the sequence N. Can we just to L now, since L is the greatest No, it bound after said of elements of sequences of the terms of sequences. So for any absolutely. Didn't see role as minus epsilon, which is strictly less than N. I'm sorry, L plus epsilon, which is strictly created. An L cannot be by lower bound. So there exists some. And in particular there exists some term A N. That is less than L plus of silence. But since in is a decreasing sequence we have a N less than equal to capital and for every and greater than or equal to capital. And so in particular we have and less than L plus a silent for every angry to an end no sense. And is can feel them. We have less than equal to a N. For every n natural numbers. In particular, l minus epsilon, which is strictly less than L, would also be less than a N for every angrier than capital. And so we have this inequality right here and this inequality right here and we combined them to find that for every epsilon created in zero, there exists a national number and such side for every end. After that comes after that natural number added minus upside and is less than a. In which is less than L. Plus of silent. That means A. In the sequence A in Canada. Just to L. This completes option for this term.

A role for this problem is to determine where the function is continuous or where it isn't continuous. So um it's important to know that in order for a function to be continuous, the limit as X approaches C of F of X is equal to F F C. So what that means is that if we have something like X squared, we know this function is continuous at X equals zero because the limit as X approaches C is zero and that's the same value as the function at sea. So we know that that is where the function will be continuous. However, if we had something like this right here, then what we end up seeing is that this function is not continuous because the limit as X approaches here is going to be one of the limit as X approaches here is going to be negative one. So the limit doesn't exist and we see the function has these different values. So that's how we make sense of continuity.

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.

For this question. We just need to rewrite the function G a little bit. So we have G X equal to the integral of F. T. From X to be FT. D. T. Um So we want to investigate this function is continuous and differentiable at a particular point in the interval A from B. So we know that F. S anti global function. So from A to B we have F T. D. T. Actually it's equal to from A to X F T. D. T. Class from X to be F T. D. T. Right? So this is the rule of domain act additive T. Um so now we can rewrite um this one uh and we know that this function is anti global. So we can have a constant here because we can integrate this function over this interval, so we will get a real value and we assume this is C. So now we can rewrite um this part using this constant and this part, so now we have G X. As equal to c minus. Um this an integral from A to X FT D. T. And we know that we assume um this one S H. X. Okay, so H X is equal to as equal to this anti grow, have T. D. T. Given this using the from the mental theorem of calculus and we know that H X is a uniform, uniform needs continuous um function. So H X is you knew for money um continue continuous interval A G. P. And differentiate trouble on the open interval of this open interval. Right? So H X is um this thing and now we can have that this is a constant minus uniformly continuous function of this um closed interval. So we know that G X. Is continuous and this interval and then we can easily get that G X. Yes continuous. Right? And then since we have this um transformation and we know that that this um derivative as equal to as um Halifax and then we know that G. Um if we want to evaluate the distributive at a particular point C we will have which is equal to negative Fc. Because um okay so to to distinguish we need to notify that actually this is a big C as different from this. Mostly given the question. So I just want to um clarify to avoid any confusion. So the derivative of this constant with with respect to X will be zero. So she prime X will be equal to negative fx. And then G prime C will be negative of state


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