Homework If the inverse T 1 of a closed linear operator exists, show that T 1 is a closed linear operator.

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1 Homework 3 1 If the inverse T 1 of a closed linear operator exists, show that T 1 is a closed linear operator Solution: Assuming that the inverse of T were defined, then we will have to have that D(T 1 ) = Ran(T ) Suppose that {u n } D(T 1 ) is a sequence such that u n u and T 1 u n x We need so that that u D(T 1 ) and T 1 u = x Then by the simple criteria, we would have that T 1 is closed Since u n D(T 1 ) we have that u n = T x n, where x n D(T ) Also, note we have that T 1 u n = x n x So, we have a sequence of vectors {x n } D(T ) with x n x and T x n u Since T is closed, we have that x D(T ) and T x = u But, this last equality implies that u D(T 1 ) and that x = T 1 u, which is what we wanted 2 Let T be a closed linear operator If two sequences {x n } and { x n } in the domain of T both converge to the same limit x, and if {T x n } and {T x n } both converge, show that {T x n } and {T x n } both have the same limit Solution: Suppose that {x n }, { x n } D(T ) are such that x n x and x n x Let T x n y and T x n ỹ We need to show that y = ỹ Since T is closed, then we know that for x n x and T x n y implies that x D(T ) and T x = y Similarly, we have that x n x and T x n ỹ implies that x D(T ) and T x = ỹ But, this then gives y = T x = ỹ, and we they have the same limit as claimed 3 Let X and Y be Banach spaces and T : X Y an injective bounded linear operator Show that T 1 : Ran T X is bounded if and only if Ran T is closed in Y Solution: First suppose that Ran T is closed in Y By problem 1 we know that T 1 is closed and D(T 1 ) = Ran(T ) Then by the Closed Graph Theorem, we have that T 1 is a bounded linear operator from Ran T X Now, suppose that T 1 : Ran T X is bounded, ie, T 1 y X c y Y The goal is to show that y Ran T Let y Ran T, and then there exists x n X such that T x n := u n y We then have that the sequence {x n } is Cauchy since x n x m X = T 1 T x n T 1 T x m X = T 1 (u n u m ) X c u n u m Y and since u n y this gives that the claim Since X is complete, let x denote the limit of the sequence {x n } Then we will show that y = T x Ran T, and so the range of T is

2 closed But, note that y T x Y = y u n + u n T x Y y u n Y + u n T x Y y u n Y + T x n x X Here we have used that T is bounded But, the expression on the right hand side can be made arbitrarily small, and so y = T x as desired 4 Let T : X Y be a bounded linear operator, where X and Y are Banach spaces If T is bijective, show that there are positive real numbers a and b such that a x X T x Y b x X x X Solution: Since T is a bounded linear operator, then we immediately have that T x Y T x X x X However, since T is bijective and bounded we have that T 1 is bounded as well Note that T 1 : Y X, so for any y Y we have that T 1 y X T 1 y Y Apply this when y = T x, and we then have Combining inequalities we have that x X T 1 T x Y a x X T x Y b x X x X with a = T 1 1 and b = T 5 Let X 1 = (X, 1 ) and X 2 = (X, 2 ) be Banach spaces If there is a constant c such that x 1 c x 2 for all x X, show that there is a constant C such that x 2 for all x X Solution: Consider the map I : X 2 X 1 given by Ix = x Then I is clearly linear from X 2 X 1, and by the hypotheses, we have Ix 1 c x 2 x X so I is bounded But, we also have that I is bijective, and so by the Bounded Inverse Theorem, we have that I 1 : X 1 X 2 is a bounded linear operator as well, ie, there

3 exists a constant C such that I 1 x 2 x X However, I 1 x = x and so x 2 x X 6 Let X be the normed space whose points are sequences of complex numbers x = {ξ j } with only finitely many non-zero terms and norm defined by x = sup j ξ j Let T : X X be defined by ( T x = ξ 1, 1 2 ξ 2, 1 ) ( ) 1 3 ξ 3, = j ξ j Show that T is linear and bounded, but T 1 is unbounded Does this contradict the Bounded Inverse Theorem? Solution: It is pretty obvious that T is linear Since x = {ξ n } and y = {η n }, then we have that { } { } { } T (x + y) = j (ξ j + η j ) = j ξ j + j η j = T x + T y The case of constants λ, is just as easy to verify, T λx = λt x It is also very easy to see that T is bounded, since we have for any x X that T x X = sup j 1 ξ j j x X sup j 1 1 j x X We now show that T 1 is not bounded First, note that T 1 is given by T 1 x = {jξ j } Now let x = {ξ j } where ξ j = 0 if j n and 1 if n = j Then we have that x X = 1 and T 1 x = {jξ j } = (0,, 0, n, 0,, ) where the value 1 occurs in the nth coordinate This then implies that T 1 x X = n But, this is enough to show that the operator is not bounded This doesn t contradict the Bounded Inverse Theorem since X is not complete 7 Give an example to show that boundedness need not imply closedness Solution: Let T : D(T ) D(T ) X be the identity operator on D(T ), where D(T ) is a proper dense subspace of the normed space X Then it is immediate that T is linear and bounded However, we have that T is not closed To see this, take x X \ D(T ) and a sequence {x n } D(T ) that converges to x, and then use the simple test for an operator to be closed that we gave in class

4 8 Show that the null space of a closed linear operator T : X Y is a closed subspace of X Solution: Recall that the null space of a linear operator is given by Null(T ) := {x X : T x = 0} Let x Null(T ) and suppose that x n Null(T ) is a sequence such that x n x We need to show that x Null(T ) If we have that T x n y, then since each x n Null(T ) we have that y = 0 But since T is closed, then we have that T x = y = 0 and so x Null(T ) 9 Let H be a Hilbert space and suppose that A : H H is everywhere defined and Show that A is a bounded linear operator Ax, y H = x, Ay H x, y H Solution: It suffices to show that the graph of A is closed So suppose that we have (x n, Ax n ) (x, y) Then we need to show that y = Ax To accomplish this last equality, it is enough to show that for all z H that z, y H = z, Ax H But, with this observation the problem becomes rather easy: z, y H = lim n z, Ax n H = lim n Az, x n H = Az, x H = z, Ax H 10 Let {T n } be a sequence of bounded linear transformations from X Y, where X and Y are Banach spaces Suppose that T n M < and there is a dense set E X such that {T n x} converges for all x E Show that {T n x} converges on all of X Solution: Let ɛ > 0 be given Pick any x X Now choose x 0 E such that x x 0 X < ɛ, which will be possible since E is dense in X For this x 3M 0 we have that T n x 0 y So there exists an integer N such that if n, m > N then T n x 0 T m x 0 Y < ɛ 3 Now, we claim that {T n x} is Cauchy in Y, and since Y is a Banach space it is complete

5 and so the result follows To see that {T n x} is Cauchy, when n, m > N we have T n x T m x Y = T n x T n x 0 + T n x 0 T m x 0 + T m x 0 T m x Y T n x T n x 0 Y + T n x 0 T m x 0 Y + T m x T m x 0 Y < T n x x 0 X + T m x x 0 X + ɛ 3 < 2M ɛ 3M + ɛ 3 = ɛ