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 fung 2 D(T ) is a sequence such that un ! u and −1 −1 −1 T un ! x. We need so that that u 2 D(T ) and T u = x. Then by the simple criteria, we would have that T −1 is closed. −1 Since un 2 D(T ) we have that un = T xn, where xn 2 D(T ). Also, note we have that −1 T un = xn ! x. So, we have a sequence of vectors fxng 2 D(T ) with xn ! x and T xn ! u. Since T is closed, we have that x 2 D(T ) and T x = u. But, this last equality implies that u 2 D(T −1) and that x = T −1u, which is what we wanted. 2. Let T be a closed linear operator. If two sequences fxng and fx~ng in the domain of T both converge to the same limit x, and if fT xng and fT x~ng both converge, show that fT xng and fT x~ng both have the same limit. Solution: Suppose that fxng; fx~ng 2 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 2 D(T ) and T x = y. Similarly, we have thatx ~n ! x and T x~n ! y~ implies that x 2 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 2 Ran T . Let y 2 Ran T , and then there exists xn 2 X such that T xn := un ! y. We then have that the sequence fxng 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 fxng. Then we will show that y = T x 2 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 8x 2 X: Solution: Since T is a bounded linear operator, then we immediately have that kT xkY ≤ kT k kxkX 8x 2 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 2 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 8x 2 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 2 X, show that there is a constant C such that kxk2 ≤ C kxk1 for all x 2 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 8x 2 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 8x 2 X: However, I−1x = x and so kxk2 ≤ C kxk1 8x 2 X: 6. Let X be the normed space whose points are sequences of complex numbers x = fξjg with only finitely many non-zero terms and norm defined by kxk = supj jξjj. 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 = fξng and y = fηng, 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 2 X that jξjj 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 = fjξjg : Now let x = fξjg where ξj = 0 if j 6= n and 1 if n = j. Then we have that kxkX = 1 −1 and T x = fjξjg = (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 2 X n D(T ) and a sequence fxng 2 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 ) := fx 2 X : T x = 0g. Let x 2 Null(T ) and suppose that xn 2 Null(T ) is a sequence such that xn ! x. We need to show that x 2 Null(T ). If we have that T xn ! y, then since each xn 2 Null(T ) we have that y = 0. But since T is closed, then we have that T x = y = 0 and so x 2 Null(T ). 9. Let H be a Hilbert space and suppose that A : H ! H is everywhere defined and hAx; yiH = hx; AyiH 8x; y 2 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 2 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 fTng be a sequence of bounded linear transformations from X ! Y , where X and Y are Banach spaces. Suppose that kTnk ≤ M < 1 and there is a dense set E ⊂ X such that fTnxg converges for all x 2 E. Show that fTnxg converges on all of X. Solution: Let > 0 be given. Pick any x 2 X. Now choose x0 2 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 fTnxg is Cauchy in Y , and since Y is a Banach space it is complete and so the result follows. To see that fTnxg 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.