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 −1 −1 D(T ) = Ran(T ). Suppose that {un} ∈ D(T ) is a sequence such that un → u and −1 −1 −1 T un → x. We need so that that u ∈ D(T ) and T u = x. Then by the simple criteria, we would have that T −1 is closed.
−1 Since un ∈ D(T ) we have that un = T xn, where xn ∈ D(T ). Also, note we have that −1 T un = xn → x. So, we have a sequence of vectors {xn} ∈ D(T ) with xn → x and T xn → 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 −1u, which is what we wanted.
2. Let T be a closed linear operator. If two sequences {xn} and {x˜n} in the domain of T both converge to the same limit x, and if {T xn} and {T x˜n} both converge, show that {T xn} and {T x˜n} both have the same limit.
Solution: Suppose that {xn}, {x˜n} ∈ D(T ) are such that xn → x andx ˜n → x. Let T xn → y and T x˜n → y˜. We need to show that y =y ˜.
Since T is closed, then we know that for xn → x and T xn → y implies that x ∈ D(T ) and T x = y. Similarly, we have thatx ˜n → x and T x˜n → y˜ implies that x ∈ D(T ) and T x =y ˜. But, this then gives y = T x =y ˜, 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.
−1 −1 Now, suppose that T : Ran T → X is bounded, i.e., kT ykX ≤ c kykY . The goal is to show that y ∈ Ran T . Let y ∈ Ran T , and then there exists xn ∈ X such that T xn := un → y. We then have that the sequence {xn} is Cauchy since
−1 −1 −1 kxn − xmkX = T T xn − T T xm X = T (un − um) X ≤ c kun − umkY
and since un → y this gives that the claim. Since X is complete, let x denote the limit of the sequence {xn}. Then we will show that y = T x ∈ Ran T , and so the range of T is closed. But, note that
ky − T xkY = ky − un + un − T xkY
≤ ky − unkY + kun − T xkY
≤ ky − unkY + kT k kxn − xkX . 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 kxkX ≤ kT xkY ≤ b kxkX ∀x ∈ X.
Solution: Since T is a bounded linear operator, then we immediately have that
kT xkY ≤ kT k kxkX ∀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
−1 −1 T y X ≤ T kykY . Apply this when y = T x, and we then have
−1 kxkX ≤ T kT xkY . Combining inequalities we have that
a kxkX ≤ kT xkY ≤ b kxkX ∀x ∈ X
with a = kT −1k−1 and b = kT k.
5. Let X1 = (X, k·k1) and X2 = (X, k·k2) be Banach spaces. If there is a constant c such that kxk1 ≤ c kxk2 for all x ∈ X, show that there is a constant C such that kxk2 ≤ C kxk1 for all x ∈ X.
Solution: Consider the map I : X2 → X1 given by Ix = x. Then I is clearly linear from X2 → X1, and by the hypotheses, we have
kIxk1 ≤ c kxk2 ∀x ∈ X so I is bounded. But, we also have that I is bijective, and so by the Bounded Inverse −1 Theorem, we have that I : X1 → X2 is a bounded linear operator as well, i.e., there exists a constant C such that
−1 I x 2 ≤ C kxk1 ∀x ∈ X. However, I−1x = x and so
kxk2 ≤ C kxk1 ∀x ∈ X.
6. Let X be the normed space whose points are sequences of complex numbers x = {ξj} with only finitely many non-zero terms and norm defined by kxk = supj |ξj|. Let T : X → X be defined by 1 1 1 T x = ξ , ξ , ξ ,... = ξ . 1 2 2 3 3 j 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 = {ξn} and y = {ηn}, then we have that 1 1 1 T (x + y) = (ξ + η ) = ξ + η = T x + T y. j j j j j j j 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
|ξj| 1 kT xkX = sup ≤ kxkX sup ≤ kxkX . j≥1 j j≥1 j
We now show that T −1 is not bounded. First, note that T −1 is given by
−1 T x = {jξj} .
Now let x = {ξj} where ξj = 0 if j 6= n and 1 if n = j. Then we have that kxkX = 1 −1 and T x = {jξj} = (0,..., 0, n, 0,..., ) where the value 1 occurs in the nth coordinate. −1 This then implies that kT xkX = 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 {xn} ∈ D(T ) that converges to x, and then use the simple test for an operator to be closed that we gave in class. 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 xn ∈ Null(T ) is a sequence such that xn → x. We need to show that x ∈ Null(T ).
If we have that T xn → y, then since each xn ∈ 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
hAx, yiH = hx, AyiH ∀x, y ∈ H. Show that A is a bounded linear operator.
Solution: It suffices to show that the graph of A is closed. So suppose that we have (xn, Axn) → (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
hz, yiH = hz, AxiH . But, with this observation the problem becomes rather easy:
hz, yi = lim hz, Axni H n H = lim hAz, xni n H
= hAz, xiH
= hz, AxiH .
10. Let {Tn} be a sequence of bounded linear transformations from X → Y , where X and Y are Banach spaces. Suppose that kTnk ≤ M < ∞ and there is a dense set E ⊂ X such that {Tnx} converges for all x ∈ E. Show that {Tnx} converges on all of X.
Solution: Let > 0 be given. Pick any x ∈ X. Now choose x0 ∈ E such that kx − x0kX < 3M , which will be possible since E is dense in X. For this x0 we have that Tnx0 → y. So there exists an integer N such that if n, m > N then kT x − T x k < . n 0 m 0 Y 3
Now, we claim that {Tnx} is Cauchy in Y , and since Y is a Banach space it is complete and so the result follows. To see that {Tnx} is Cauchy, when n, m > N we have
kTnx − TmxkY = kTnx − Tnx0 + Tnx0 − Tmx0 + Tmx0 − TmxkY
≤ kTnx − Tnx0kY + kTnx0 − Tmx0kY + kTmx − Tmx0kY < kT k kx − x k + kT k kx − x k + n 0 X m 0 X 3 < 2M + = . 3M 3