Differentiation is linear: the derivative of a sum is the sum of the derivatives, and constants factor out. On the space of polynomials, with size measured by the largest absolute value on \([0,1]\) , the monomial \(x^n\) has size \(1\) and its derivative \(nx^{n-1}\) has size \(n\) . A unit vector is sent to a vector of size \(n\) , for every \(n\) . No matrix does this: a linear map on \(\mathbb R^d\) stretches unit vectors by at most a fixed factor. So differentiation, on this space with this norm, is a linear map that is not continuous, since polynomials \(x^n/n\) tending to zero have derivatives that do not.

The lesson is that in infinite dimensions linearity and continuity are separate properties, and the maps worth studying are those that have both. This chapter shows that continuity of a linear map is equivalent to a single inequality, turns the best constant in that inequality into a norm, and computes that norm for the operators that recur in the rest of the book: matrices, shifts, multiplications, and integral operators. It ends with two tools that depend on completeness, extension from a dense subspace and the Neumann series for inverting a perturbation of the identity.

2.1 Boundedness is continuity

A linear map between normed spaces is usually called an operator; when the target is the scalar field it is a functional. We write \(Tx\) for \(T(x)\) .

The word bounded is used in a special sense here. A nonzero linear map is never a bounded function on the whole of \(X\) ; what is bounded is its restriction to the unit ball. Indeed the smallest admissible constant is

the three suprema being equal by homogeneity (Exercise 2). Two inequalities are used constantly: \(\lVert Tx\rVert\le\lVert T\rVert\,\lVert x\rVert\) for every \(x\) , and, for bounded \(T:X\to Y\) and \(S:Y\to Z\) ,

which follows from \(\lVert STx\rVert\le\lVert S\rVert\lVert Tx\rVert\le\lVert S\rVert\lVert T\rVert\lVert x\rVert\) .

Linearity converts a local, qualitative property into a global, quantitative one. By Exercise 17 of Chapter 1Chapter 1 every linear map defined on a finite-dimensional normed space is bounded, so the distinction is invisible in linear algebra.

2.2 Computing a norm from both sides

To prove \(\lVert T\rVert=c\) one must do two different things. An upper bound \(\lVert Tx\rVert\le c\lVert x\rVert\) is an inequality valid for every \(x\) . A lower bound is a single unit vector, or a sequence of unit vectors, on which \(\lVert Tx\rVert\) equals or approaches \(c\) . Neither half can be omitted, and they call on different skills: the first is an estimate, the second a construction.

For matrices with the \(1\) -norm or the \(\infty\) -norm both halves can be carried out in general.

When both spaces carry the Euclidean norm, \(\lVert A\rVert^2\) is the largest eigenvalue of the Hermitian matrix \(A^*A\) , and \(\lVert A\rVert\) is the largest singular value of \(A\) . This is a theorem of Linear Algebra, resting on the orthogonal diagonalisation of Hermitian matrices; Chapter 10 extends it to compact operators.

2.3 Operators on sequences and functions

Four families of operators supply most of the examples in this subject. For each we find the norm by the two-sided method.

2.4 The space of bounded operators

The set \(\mathcal B(X,Y)\) is a vector space under pointwise operations, and the operator norm is a norm on it: definiteness and homogeneity are immediate, and \(\lVert(S+T)x\rVert\le\lVert Sx\rVert+\lVert Tx\rVert\le(\lVert S\rVert+\lVert T\rVert)\lVert x\rVert\) gives the triangle inequality.

The structure of this proof repeats that of Theorem 1.3: a pointwise limit is found first, and a uniform estimate promotes it to a limit in norm. Only completeness of the target was used. The domain may be any normed space, which is why duals are always complete, even the dual of \(c_{00}\) .

A bounded operator \(T:X\to Y\) is called invertible if it is bijective and its inverse is bounded; an invertible operator is also called an isomorphism of normed spaces, and it is an isometry if moreover \(\lVert Tx\rVert=\lVert x\rVert\) for all \(x\) . The kernel \(\ker T=\{x:Tx=0\}\) of a bounded operator is always a closed subspace, being the preimage of the closed set \(\{0\}\) under a continuous map. The range \(\operatorname{ran}T=T(X)\) is a subspace that need not be closed, and Chapter 5 will show that whether it is closed governs whether the inverse is bounded.

2.5 Extension from a dense subspace

Operators are often defined first on a convenient dense subspace, such as polynomials, step functions, or finitely supported sequences, where a formula makes obvious sense. Boundedness lets the definition spread to the whole space.

Both hypotheses are needed. Completeness of \(Y\) is what makes the Cauchy sequence \((T_0d_n)\) converge. Boundedness is what makes it Cauchy: the evaluation functional \(f\mapsto f(0)\) is defined on the dense subspace \(C[0,1]\) of \(L^1[0,1]\) , but it is not bounded for \(\lVert\cdot\rVert_1\) (Exercise 15), and indeed “the value at \(0\) ” of an element of \(L^1\) has no meaning.

2.6 Inverting a small perturbation of the identity

For a real number \(t\) with \(\lvert t\rvert<1\) , the geometric series gives \((1-t)^{-1}=\sum_{n\ge 0}t^n\) . In a Banach space the same formula inverts operators.

2.7 Two ways for operators to converge

Convergence in norm implies strong convergence, since \(\lVert T_nx-Tx\rVert\le\lVert T_n-T\rVert\lVert x\rVert\) ; it is uniform convergence on the unit ball, whereas strong convergence is pointwise convergence. In finite dimensions the two coincide. In infinite dimensions the gap between them is the gap between a ball and a compact set.

Exercises

1. Compute the operator norm of \(A=\begin{pmatrix}1&1\\0&1\end{pmatrix}\) on \(\mathbb R^2\) with respect to \(\lVert\cdot\rVert_1\) , \(\lVert\cdot\rVert_\infty\) , and \(\lVert\cdot\rVert_2\) , and give a unit vector attaining the norm in each case.

   Answer

   For \(\lVert\cdot\rVert_1\) : column sums \(1\) and \(2\) , so \(\lVert A\rVert=2\) , attained at \(e_2\) . For \(\lVert\cdot\rVert_\infty\) : row sums \(2\) and \(1\) , so \(\lVert A\rVert=2\) , attained at \((1,1)\) . For \(\lVert\cdot\rVert_2\) : \(A^{\top}A=\begin{pmatrix}1&1\\1&2\end{pmatrix}\) has largest eigenvalue \((3+\sqrt5)/2\) , whose square root is the golden ratio, so \(\lVert A\rVert=(1+\sqrt5)/2\approx 1.618\) , attained at the unit vector proportional to \((1,(1+\sqrt5)/2)\) .

2. Let \(T:X\to Y\) be a bounded operator with \(X\ne\{0\}\) . Prove that \(\sup_{x\ne 0}\lVert Tx\rVert/\lVert x\rVert\) , \(\sup_{\lVert x\rVert\le 1}\lVert Tx\rVert\) , and \(\sup_{\lVert x\rVert=1}\lVert Tx\rVert\) are equal and equal the smallest constant \(C\) in Definition 2.1.

   Hint

   Call the three suprema \(a\) , \(b\) , \(c\) in the order listed. Homogeneity gives \(\lVert Tx\rVert/\lVert x\rVert=\lVert T(x/\lVert x\rVert)\rVert\) , so \(a=c\) ; clearly \(c\le b\) ; and for \(0<\lVert x\rVert\le 1\) , \(\lVert Tx\rVert\le\lVert Tx\rVert/\lVert x\rVert\le a\) , so \(b\le a\) . Finally a constant \(C\) is admissible in Definition 2.1 exactly when \(C\ge a\) .

3. Let \((d_k)\) be a bounded scalar sequence and \((Dx)_k=d_kx_k\) on \(\ell^2\) . Prove that \(\lVert D\rVert=\sup_k\lvert d_k\rvert\) , and that the norm is attained at some unit vector if and only if the supremum is attained at some index.

   Solution

   Let \(d=\sup_k\lvert d_k\rvert\) . For every \(x\in\ell^2\) ,

   \[ \lVert Dx\rVert_2^2=\sum_k\lvert d_k\rvert^2\lvert x_k\rvert^2\le d^2\sum_k\lvert x_k\rvert^2=d^2\lVert x\rVert_2^2 , \]

   so \(D\) maps \(\ell^2\) to itself and \(\lVert D\rVert\le d\) . For the lower bound, \(\lVert De_k\rVert_2=\lvert d_k\rvert\) with \(\lVert e_k\rVert_2=1\) , so \(\lVert D\rVert\ge\lvert d_k\rvert\) for every \(k\) , hence \(\lVert D\rVert\ge d\) . Thus \(\lVert D\rVert=d\) .

   If the supremum is attained, say \(\lvert d_j\rvert=d\) , then \(e_j\) is a unit vector with \(\lVert De_j\rVert_2=d\) . Conversely suppose \(\lVert Dx\rVert_2=d\) for some unit vector \(x\) . Subtracting,

   \[ 0=d^2\lVert x\rVert_2^2-\lVert Dx\rVert_2^2=\sum_k\bigl(d^2-\lvert d_k\rvert^2\bigr)\lvert x_k\rvert^2 . \]

   Every term is nonnegative, so every term vanishes. Since \(x\ne 0\) there is an index \(k\) with \(x_k\ne 0\) , and for it \(\lvert d_k\rvert=d\) : the supremum is attained. Example 2.3 is the case \(d_k=1-1/k\) , where the supremum \(1\) is not attained by any \(d_k\) .

4. Let \(\varphi\in C[0,1]\) and \(M_\varphi f=\varphi f\) on \(C[0,1]\) with the supremum norm. Prove that \(\lVert M_\varphi\rVert=\lVert\varphi\rVert_\infty\) and that the norm is always attained.

   Hint

   The upper bound is \(\lvert\varphi(x)f(x)\rvert\le\lVert\varphi\rVert_\infty\lVert f\rVert_\infty\) . For the lower bound test on the constant function \(1\) , which belongs to \(C[0,1]\) and has norm \(1\) : \(M_\varphi 1=\varphi\) . The contrast with Example 2.3 is that \(\ell^2\) contains no constant sequence other than \(0\) .

5. Find the norms of the functionals \(f\mapsto f(\tfrac12)\) and \(f\mapsto f(0)-f(1)\) on \(C[0,1]\) with the supremum norm, exhibiting in each case a function of norm \(1\) at which the norm is attained.

   Answer

   The functional \(f\mapsto f(\tfrac12)\) has norm \(1\) , attained at the constant function \(1\) . The functional \(f\mapsto f(0)-f(1)\) satisfies \(\lvert f(0)-f(1)\rvert\le 2\lVert f\rVert_\infty\) and takes the value \(2\) at \(f(x)=1-2x\) , which has norm \(1\) ; its norm is \(2\) .

6. Prove that the Volterra operator satisfies \((V^nf)(x)=\int_0^x\frac{(x-t)^{n-1}}{(n-1)!}f(t)\,dt\) for \(n\ge 1\) , and deduce again that \(\lVert V^n\rVert=1/n!\) on \(C[0,1]\) .

   Solution

   Write \((W_nf)(x)=\int_0^x\frac{(x-t)^{n-1}}{(n-1)!}f(t)\,dt\) . For \(n=1\) this is \(Vf\) . Suppose \(V^n=W_n\) . Then

   \[ (V^{n+1}f)(x)=\int_0^x(W_nf)(s)\,ds=\int_0^x\int_0^s\frac{(s-t)^{n-1}}{(n-1)!}f(t)\,dt\,ds . \]

   The integrand is continuous on the triangle \(0\le t\le s\le x\) , so the order of integration may be exchanged (math.multivariable-calculus.fubini-continuous): for fixed \(t\) the variable \(s\) runs from \(t\) to \(x\) , and

   \[ (V^{n+1}f)(x)=\int_0^xf(t)\int_t^x\frac{(s-t)^{n-1}}{(n-1)!}\,ds\,dt=\int_0^x\frac{(x-t)^{n}}{n!}f(t)\,dt=(W_{n+1}f)(x). \]

   This completes the induction. For the norm, the kernel is nonnegative, so

   \[ \lvert V^nf(x)\rvert\le\lVert f\rVert_\infty\int_0^x\frac{(x-t)^{n-1}}{(n-1)!}\,dt=\frac{x^n}{n!}\lVert f\rVert_\infty\le\frac{1}{n!}\lVert f\rVert_\infty , \]

   which gives \(\lVert V^n\rVert\le 1/n!\) , and \(f=1\) gives \(V^n1=x^n/n!\) with supremum \(1/n!\) . Hence \(\lVert V^n\rVert=1/n!\) .

7. Let \((Kf)(x)=\int_0^1xy\,f(y)\,dy\) . Compute \(\lVert K\rVert\) when \(K\) acts on \(C[0,1]\) with the supremum norm, and again when \(K\) acts on \(L^2[0,1]\) , and explain why the two answers differ.

   Solution

   On \(C[0,1]\) . The kernel \(k(x,y)=xy\) is continuous and nonnegative, so Proposition 2.4 gives

   \[ \lVert K\rVert=\max_{0\le x\le 1}\int_0^1xy\,dy=\max_{0\le x\le 1}\frac{x}{2}=\frac12 , \]

   attained at the constant function \(1\) , for which \(K1=x/2\) .

   On \(L^2[0,1]\) . Write \(u(x)=x\) . Then \(Kf=\langle f,u\rangle\,u\) , where \(\langle f,u\rangle=\int_0^1yf(y)\,dy\) . By the Cauchy–Schwarz inequality,

   \[ \lVert Kf\rVert_2=\lvert\langle f,u\rangle\rvert\,\lVert u\rVert_2\le\lVert u\rVert_2^2\,\lVert f\rVert_2=\tfrac13\lVert f\rVert_2 , \]

   because \(\lVert u\rVert_2^2=\int_0^1x^2\,dx=\tfrac13\) . Equality holds for \(f=u/\lVert u\rVert_2=\sqrt3\,x\) . So \(\lVert K\rVert=\tfrac13\) on \(L^2[0,1]\) .

   The answers differ because the operator norm depends on the norms of the spaces, not only on the formula. The extremal function for the supremum norm is the constant \(1\) ; in \(L^2\) the constant \(1\) also has norm \(1\) but gives only \(\lVert K1\rVert_2=\lVert x/2\rVert_2=1/(2\sqrt3)\approx 0.289\) , less than \(\tfrac13\) , and the best direction is \(f\) proportional to \(y\) .

8. Show that on \(\ell^p\) , \(1\le p\le\infty\) , the right shift is an isometry that is not surjective and the left shift is surjective with \(\lVert L\rVert=1\) and one-dimensional kernel; then explain which theorem of linear algebra forbids such a pair of square matrices.

   Hint

   For \(p<\infty\) , \(\lVert Sx\rVert_p^p=\sum_k\lvert x_k\rvert^p\) and \(\lVert Lx\rVert_p^p=\lVert x\rVert_p^p-\lvert x_1\rvert^p\) ; for \(p=\infty\) use suprema. \(\ker L\) is spanned by \(e_1\) , and \(L(Sx)=x\) shows \(L\) is onto. For square matrices, the rank and nullity count shows that injective, surjective, and bijective are equivalent, so no matrix can be injective without being surjective.

9. On \(c_{00}\) with the norm \(\lVert\cdot\rVert_2\) define \(\sigma(x)=\sum_kx_k\) . A student argues: “\(\lvert\sigma(e_k)\rvert=1\) for every basis vector, so \(\lVert\sigma\rVert=1\) .” Show that \(\sigma\) is in fact unbounded, and state precisely what the student’s computation does prove.

   Solution

   Let \(u^{(n)}=n^{-1/2}(1,\dots,1,0,0,\dots)\) with \(n\) ones. Then \(u^{(n)}\in c_{00}\) , \(\lVert u^{(n)}\rVert_2=n^{-1/2}\sqrt n=1\) , and

   \[ \sigma(u^{(n)})=n\cdot n^{-1/2}=\sqrt n . \]

   So \(\sup_{\lVert x\rVert_2=1}\lvert\sigma(x)\rvert\ge\sqrt n\) for every \(n\) , and \(\sigma\) is unbounded. What the student computed is that \(\lvert\sigma(x)\rvert=1\) at the particular unit vectors \(x=e_k\) . That proves the lower bound \(\lVert\sigma\rVert\ge 1\) and nothing more. An upper bound requires an inequality \(\lvert\sigma(x)\rvert\le C\lVert x\rVert_2\) for all \(x\) , and none exists. The underlying misconception is that a linear map is controlled by its values on a basis; that is true algebraically, since those values determine the map, but the norm of a combination \(\sum_ka_ke_k\) is \((\sum_k\lvert a_k\rvert^2)^{1/2}\) , which can be much smaller than \(\sum_k\lvert a_k\rvert\) , and the functional adds the coefficients without any such saving.

10. Let \(f\) be a linear functional on a normed space \(X\) . Prove that \(f\) is bounded if and only if \(\ker f\) is closed.

    Solution

    If \(f\) is bounded it is continuous by Theorem 2.2, and \(\ker f\) is the preimage of the closed set \(\{0\}\) , hence closed.

    For the converse we prove the contrapositive. Suppose \(f\) is not bounded; in particular \(f\ne 0\) , so there is \(x_0\) with \(f(x_0)=1\) . Unboundedness gives, for each \(n\) , a vector \(x_n\) with \(\lVert x_n\rVert\le 1\) and \(\lvert f(x_n)\rvert\ge n\) . Put

    \[ z_n=x_0-\frac{x_n}{f(x_n)} . \]

    Then \(f(z_n)=1-1=0\) , so \(z_n\in\ker f\) , and \(\lVert z_n-x_0\rVert=\lVert x_n\rVert/\lvert f(x_n)\rvert\le 1/n\to 0\) . Thus \(x_0\) is a limit of elements of \(\ker f\) , but \(x_0\notin\ker f\) since \(f(x_0)=1\) . So \(\ker f\) is not closed. The argument shows more: every \(x_0\) with \(f(x_0)=1\) , and hence by scaling every vector, lies in the closure of \(\ker f\) . The kernel of an unbounded functional is dense.

11. Let \(\mathcal P_1\) be the polynomials of degree at most \(1\) on \([0,1]\) with the supremum norm. Show that the derivative \(D:\mathcal P_1\to\mathcal P_1\) has \(\lVert D\rVert=2\) and find a polynomial attaining it.

    Answer

    For \(p(x)=a+bx\) we have \(\lVert p\rVert_\infty=\max\{\lvert a\rvert,\lvert a+b\rvert\}\) and \(\lVert Dp\rVert_\infty=\lvert b\rvert\le\lvert a\rvert+\lvert a+b\rvert\le 2\lVert p\rVert_\infty\) . Equality holds for \(p(x)=1-2x\) , which has \(\lVert p\rVert_\infty=1\) and \(\lVert Dp\rVert_\infty=2\) . So \(\lVert D\rVert=2\) .

12. On \(\ell^2\) let \(T=\tfrac12S\) , where \(S\) is the right shift. Use the Neumann series to compute \(x=(I-T)^{-1}e_1\) , find \(\lVert x\rVert_2\) , and compare it with the bound of Theorem 2.7.

    Solution

    Since \(\lVert T\rVert=\tfrac12\lVert S\rVert=\tfrac12<1\) , Theorem 2.7 gives \((I-T)^{-1}=\sum_{n\ge 0}T^n\) with \(\lVert(I-T)^{-1}\rVert\le(1-\tfrac12)^{-1}=2\) . Now \(T^ne_1=2^{-n}S^ne_1=2^{-n}e_{n+1}\) , so

    \[ x=(I-T)^{-1}e_1=\sum_{n=0}^\infty2^{-n}e_{n+1}=\bigl(1,\tfrac12,\tfrac14,\tfrac18,\dots\bigr). \]

    Check: \(Tx=\tfrac12Sx=(0,\tfrac12,\tfrac14,\dots)\) , so \(x-Tx=(1,0,0,\dots)=e_1\) . Its norm is

    \[ \lVert x\rVert_2=\Bigl(\sum_{n=0}^\infty4^{-n}\Bigr)^{1/2}=\sqrt{\tfrac43}=\frac{2}{\sqrt3}\approx 1.1547 . \]

    The theorem’s bound is \(\lVert x\rVert_2\le 2\lVert e_1\rVert_2=2\) , which holds with room to spare. The bound \(2\) is sharp for the operator all the same: applying \((I-T)^{-1}\) to unit vectors close to multiples of \((1,1,\dots,1,0,\dots)\) produces norms approaching \(2\) , because the terms \(T^n\) then reinforce one another, whereas for \(e_1\) they land in orthogonal coordinates.

13. Let \(X\) be a Banach space and \(T\in\mathcal B(X)\) invertible. Prove that if \(\lVert S-T\rVert\le\tfrac12\lVert T^{-1}\rVert^{-1}\) then \(\lVert S^{-1}\rVert\le 2\lVert T^{-1}\rVert\) and \(\lVert S^{-1}-T^{-1}\rVert\le 2\lVert T^{-1}\rVert^2\lVert S-T\rVert\) , so that inversion is continuous.

    Hint

    Put \(E=T^{-1}(T-S)\) , so \(S=T(I-E)\) with \(\lVert E\rVert\le\tfrac12\) . Then \(S^{-1}=(I-E)^{-1}T^{-1}\) and Theorem 2.7 bounds \(\lVert(I-E)^{-1}\rVert\) by \(2\) . For the second estimate use the identity \(S^{-1}-T^{-1}=S^{-1}(T-S)T^{-1}\) .

14. Show that the powers \(L^n\) of the left shift on \(\ell^2\) converge strongly to \(0\) while \(\lVert L^n\rVert=1\) for all \(n\) , and that the powers \(S^n\) of the right shift do not converge strongly to any operator.

    Solution

    For \(x\in\ell^2\) , \(L^nx=(x_{n+1},x_{n+2},\dots)\) , so

    \[ \lVert L^nx\rVert_2^2=\sum_{k>n}\lvert x_k\rvert^2 , \]

    the tail of a convergent series, which tends to \(0\) . Hence \(L^n\to 0\) strongly. On the other hand \(\lVert L^n\rVert\le\lVert L\rVert^n=1\) and \(L^ne_{n+1}=e_1\) , so \(\lVert L^n\rVert=1\) for every \(n\) ; the convergence is not in norm.

    For the right shift, suppose \(S^n\to R\) strongly for some operator \(R\) . Then in particular \(S^ne_1=e_{n+1}\) would converge in \(\ell^2\) . But \(\lVert e_{n+1}-e_{m+1}\rVert_2=\sqrt2\) for \(n\ne m\) , so the sequence \((e_{n+1})\) is not Cauchy and cannot converge. Hence \((S^n)\) has no strong limit. The two shifts behave differently because \(L^n\) discards coordinates, pushing the mass of each fixed vector out of the space, whereas \(S^n\) is an isometry that carries each vector, undiminished, off toward coordinates of ever higher index.

15. Show that \(f\mapsto\int_0^1f\) is a bounded functional of norm \(1\) on \((C[0,1],\lVert\cdot\rVert_1)\) , and that \(f\mapsto f(0)\) is unbounded on the same space by testing it on \(h_n(x)=\max\{0,1-nx\}\) . What does Theorem 2.6 say about each?

    Hint

    \(\lvert\int_0^1f\rvert\le\lVert f\rVert_1\) with equality for \(f=1\) . For the evaluation, \(h_n(0)=1\) while \(\lVert h_n\rVert_1=1/(2n)\) . Theorem 2.6 extends the integral uniquely to a functional of norm \(1\) on the completion \(L^1[0,1]\) , and gives nothing for evaluation at \(0\) because the boundedness hypothesis fails.

16. Let \(Y\) be a normed space that is not complete and \(X\ne\{0\}\) a normed space on which a nonzero bounded functional \(\varphi\) exists. Show that \(\mathcal B(X,Y)\) is not complete, using operators of the form \(x\mapsto\varphi(x)y_n\) .

    Solution

    Since \(Y\) is not complete there is a Cauchy sequence \((y_n)\) in \(Y\) with no limit in \(Y\) . Define \(T_n:X\to Y\) by \(T_nx=\varphi(x)y_n\) . Each \(T_n\) is linear, and \(\lVert T_nx\rVert=\lvert\varphi(x)\rvert\lVert y_n\rVert\le\lVert\varphi\rVert\lVert y_n\rVert\lVert x\rVert\) , so \(T_n\in\mathcal B(X,Y)\) . In the same way

    \[ \lVert T_n-T_m\rVert=\sup_{\lVert x\rVert\le 1}\lvert\varphi(x)\rvert\,\lVert y_n-y_m\rVert=\lVert\varphi\rVert\,\lVert y_n-y_m\rVert , \]

    which tends to \(0\) as \(n,m\to\infty\) . So \((T_n)\) is a Cauchy sequence in \(\mathcal B(X,Y)\) .

    Suppose it converged in operator norm to some \(T\in\mathcal B(X,Y)\) . Because \(\varphi\ne 0\) there is \(x_0\in X\) with \(\varphi(x_0)=1\) . Convergence in norm implies convergence at each vector, so \(y_n=T_nx_0\to Tx_0\) , and \(Tx_0\) is an element of \(Y\) . That contradicts the choice of \((y_n)\) . Hence \((T_n)\) has no limit and \(\mathcal B(X,Y)\) is not complete. Together with Theorem 2.5 this shows that, whenever \(X\) carries a nonzero bounded functional, \(\mathcal B(X,Y)\) is complete exactly when \(Y\) is; by the Hahn–Banach theorem of Chapter 3 every nonzero normed space carries such a functional.

17. Let \(T:X\to Y\) be an isomorphism of normed spaces. Prove that \(X\) is complete if and only if \(Y\) is, and that \(\lVert x\rVert'=\lVert Tx\rVert\) is a norm on \(X\) equivalent to the original one.

    Hint

    If \((y_n)\) is Cauchy in \(Y\) then \((T^{-1}y_n)\) is Cauchy in \(X\) because \(T^{-1}\) is bounded; apply \(T\) to its limit. For the norms, \(\lVert Tx\rVert\le\lVert T\rVert\lVert x\rVert\) and \(\lVert x\rVert=\lVert T^{-1}Tx\rVert\le\lVert T^{-1}\rVert\lVert Tx\rVert\) .

18. Let \((Tf)(x)=xf(x)+\int_0^xf(y)\,dy\) on \(C[0,1]\) with the supremum norm. Find \(\lVert T\rVert\) exactly, justifying both the upper and the lower bound.

    Answer

    \(\lVert T\rVert=2\) . Upper bound: \(\lvert Tf(x)\rvert\le x\lVert f\rVert_\infty+x\lVert f\rVert_\infty\le 2\lVert f\rVert_\infty\) . Lower bound: \(T1=2x\) , with supremum \(2\) .

19. Solve \(f(x)-\int_0^xf(y)\,dy=x\) in \(C[0,1]\) by summing the Neumann series \(\sum_nV^ng\) for \(g(x)=x\) , and verify the answer by substitution.

    Solution

    By Exercise 6 with \(f=g\) , or by direct integration, \(Vg=x^2/2\) , \(V^2g=x^3/6\) , and in general \(V^ng=x^{n+1}/(n+1)!\) : if \(V^ng=x^{n+1}/(n+1)!\) then \(V^{n+1}g=\int_0^xy^{n+1}/(n+1)!\,dy=x^{n+2}/(n+2)!\) . Theorem 2.7 applies because \(\sum_n\lVert V^n\rVert=e<\infty\) , and the unique solution is

    \[ f(x)=\sum_{n=0}^{\infty}\frac{x^{n+1}}{(n+1)!}=e^x-1 . \]

    Verification: \(\int_0^x(e^y-1)\,dy=e^x-1-x\) , so \(f(x)-\int_0^xf(y)\,dy=(e^x-1)-(e^x-1-x)=x\) , as required. The series converges uniformly on \([0,1]\) , which is what convergence in \(C[0,1]\) means, and \(\lVert f\rVert_\infty=e-1\approx 1.718\) is consistent with the bound \(\lVert f\rVert_\infty\le e\,\lVert g\rVert_\infty=e\) . Differentiating the equation gives the initial value problem \(f'-f=1\) , \(f(0)=0\) , whose solution by an integrating factor is again \(e^x-1\) ; the Neumann series has reproduced the Taylor series of that solution term by term.