0
Every uniformly convergent sequence of bounded function is uniformly bounded?
Wiki User
∙ 14y ago
The answer is yes is and only if da limit of the sequence is a bounded function.The suficiency derives directly from the definition of the uniform convergence. The necesity follows from making n tend to infinity in |fn(x)|
Wiki User
∙ 14y ago
Add your answer:
Earn +20 pts
Is is possible to have a bounded feasible region that does not optimize an objective function?
NO
How do you prove that if a real sequence is bounded and monotone it converges?
We prove that if an increasing sequence {an} is bounded above, then it is convergent and the limit is the sup {an }Now we use the least upper bound property of real numbers to say that sup {an } exists and we call it something, say S. We can say this because sup {an } is not empty and by our assumption is it bounded above so it has a LUB.Now for all natural numbers N we look at aN such that for all E, or epsilon greater than 0, we have aN > S-epsilon. This must be true, because if it were not the that number would be an upper bound which contradicts that S is the least upper bound.Now since {an} is increasing for all n greater than N we have |S-an|
What is an example of a function that is of bounded variation but not Lipschitz continuous?
How about f(x) = floor(x)? (On, say, [0,1].) It's monotone and therefore of bounded variation, but is not Lipschitz continuous (or even continuous).
Is the standard square root function bounded?
If the range is the real numbers, it has a lower bound (zero) but no upper bound.
What point is infinity?
Infinity, by definition, is not a point. Infinite means unending, so it cannot be located at a point. The moment at which a function reaches infinity is when that function ceases to be bounded.
Examples of bounded and not convergent sequence?
(0,1,0,1,...)
What is Example of bounded sequence which is not Cauchy sequence?
((-1)^n)
Every convergent sequence is bounded is the converse is true?
For the statement "convergence implies boundedness," the converse statement would be "boundedness implies convergence."So, we are asking if "boundedness implies convergence" is a true statement.Pf//By way of contradiction, "boundedness implies convergence" is false.Let the sequence (Xn) be defined asXn = 1 if n is even andXn = 0 if n is odd.So, (Xn) = {X1,X2,X3,X4,X5,X6...} = {0,1,0,1,0,1,...}Note that this is a divergent sequence.Also note that for all n, -1 < Xn < 2Therefore, the sequence (Xn) is bounded above by 2 and below by -1.As we can see, we have a bounded function that is divergent. Therefore, by way of contradiction, we have proven the converse false.Q.E.D.
Show that convergent sequence is bounded?
Wrong answer above. A limit is not the same thing as a limit point. A limit of a sequence is a limit point but not vice versa. Every bounded sequence does have at least one limit point. This is one of the versions of the Bolzano-Weierstrass theorem for sequences. The sequence {(-1)^n} actually has two limit points, -1 and 1, but no limit.
Are the major lithospheric plates bounded by diverging and converging boundaries?
They are divided by divergent, convergent AND transform boundaries.
Is is possible to have a bounded feasible region that does not optimize an objective function?
NO
How do you prove that if a real sequence is bounded and monotone it converges?
We prove that if an increasing sequence {an} is bounded above, then it is convergent and the limit is the sup {an }Now we use the least upper bound property of real numbers to say that sup {an } exists and we call it something, say S. We can say this because sup {an } is not empty and by our assumption is it bounded above so it has a LUB.Now for all natural numbers N we look at aN such that for all E, or epsilon greater than 0, we have aN > S-epsilon. This must be true, because if it were not the that number would be an upper bound which contradicts that S is the least upper bound.Now since {an} is increasing for all n greater than N we have |S-an|
Is every continuous function is bounded in C?
No. y = 1/x is continuous but unbounded.
What is the area bounded by the graph of the function fx equals 1 - e to the power of -x over the interval -1 2?
What is the area bounded by the graph of the function f(x)=1-e^-x over the interval [-1, 2]?
What sequence of events accounts for a mountain belt that is bounded on either side by cratons?
Continent-continent convergenceThe ice age and the subsequent sliding and melting of glaciers account for the mountain belt bounded on either side by craters
What sequence of events accounts for a mountain belt that is bounded of either side of cratons?
Continent-continent convergenceThe ice age and the subsequent sliding and melting of glaciers account for the mountain belt bounded on either side by craters
What is an example of a function that is of bounded variation but not Lipschitz continuous?
How about f(x) = floor(x)? (On, say, [0,1].) It's monotone and therefore of bounded variation, but is not Lipschitz continuous (or even continuous).
