Dear virtual lecturers,

just some minor comments.

I would call the “surjective isometry” in Corollary 8.1.3 an “isometric isomorphism”; in particular the latter includes the isomorphism of the linear structure.

Am I correct that the operator L is defined in Corollary 8.1.3? Maybe one should inform the reader that this is the Laplace transformation?

The notion `autonomous' plays a major role in the second part of the lecture. Maybe it deserves to be defined in the text, and not only in an exercise.

I am not particularly excited about the notation R≥0 etc., in particular if it occurs in the index in such quantity as on p.100. I completely agree that this is personal taste, but for reading I would prefer [0,∞), and similarly for other intervals in R.

Best wishes, Jürgen

Dear virtual lecturers,

I would like to share some comments from our meeting in Darmstadt.

1. On page 94 in the first section the text mentions a 'result which characterises functions in L2 with support contained in the non-negative reals'. We claim it is reasonable to say instead that the result characterises the Laplace transform of functions in L2 with support contained in the non-negative reals.
2. In Lemma 8.1.1 is notational confusion. The first equation of its proof suggests that the function f is zero almost everywhere in the negative reals. But it is not clear that this should hold by the current notation. Either you say that this is contained in the definition of L2(\R≥0;H) or you say it explicitly. In any case a further comment is necessary.
3. On page 95 in the last equation of the same proof on the right-hand it has to be C2 instead of C.

Best regards

Team Darmstadt

(How do you write the symbol for the real numbers here?)

Dear all,

in Exercise 8.5 there is a small misprint. The distribution u should be defined by u(ψ):=⟨L∗νh(i⋅+ν),(∂∗t,ν)kψ⟩L2,ν. Note that the k in the exponent of the adjoint of the time derivative is missing in the lecture notes.

Best regards

Sascha

Dear all,

I'd like to offer an alternative proof of Theorem 8.2.4 that works without Lemma 8.2.2. I do not claim that my proof is simpler, but I find it more direct because I start with the definition of M. To simplify notation, I use an alternative definition of L that omits the factor 1/√2π1/√2π; this doesn't change anything in the assertion of the theorem.

The idea is as follows: if a mapping M as asserted exists, then LTf(z)=M(z)Lf(z) holds for f=fx:=1(0,1)x, where x∈H. For this function fx I compute Lfx(z)=∫10e−ztxdt=1−e−zzx. Plugging this into LTfx(z)=M(z)Lfx(z), we see that M must be given by M(z)x:=z1−e−zLTfx(z)(x∈H).

It is readily seen that (1) defines a holomorphic function M:CRe>0→L(H). We have to show that M is bounded and satisfies the asserted equality for all f∈L2(R≥0;H) and not just for f=fx. To prove the latter, I use the fact that the set {1(a,a+1/n)x;a≥0, n∈N, x∈H} is total in L2(R≥0;H), which is why we only have to show that (LT(1(a,a+1/n)x))(z)=M(z)(L(1(a,a+1/n)x))(z) for all a≥0, n∈N, x∈H, z∈CRe>0.

For the proof of (2) observe that T(1(a,a+1/n)x)=T(1(0,1/n)x(⋅−a))=T(1(0,1/n)x)(⋅−a) since T is autonomous. Moreover, by a straightforward computation, L(f(⋅−a))(z)=e−zaLf(z) for all f∈L2(R≥0;H); in particular, (LT(1(a,a+1/n)x))(z)=e−za(LT(1(0,1/n)x))(z). Thus one sees that it suffices to show (2) in the case a=0. Also, LTfx(z)=n−1∑k=0(LT(1(k/n,(k+1)/n)x))(z)=n−1∑k=0e−zk/n(LT(1(0,1/n)x))(z)=1−e−z1−e−z/n(LT(1(0,1/n)x))(z), which by (1) and the identity (L(1(0,1/n)x))(z)=(1−e−z/n)x/z implies that (LT(1(0,1/n)x))(z)=1−e−z/nzM(z)x=M(z)(L(1(0,1/n)x))(z). This concludes the proof of (2).

EDIT: Finally we show that M is bounded. Let z∈CRe>0 and x∈H. Let f:=1R≥0e−ˉz⋅x and c:=2Rez. Then Lf(z)=∫∞0e−zt−ˉztxdt=x/c. Since LTf(z)=M(z)Lf(z) (as shown above), we obtain M(z)x=cLTf(z) and hence ‖M(z)x‖≤c∫∞0‖e−ztTf(t)‖dt≤c‖1R≥0e−z⋅‖L2(R)‖Tf‖L2(R;H)≤c‖1R≥0e−z⋅‖2L2(R)‖T‖‖x‖=‖T‖‖x‖ (note that ‖f‖L2(R;H)=‖1R≥0e−z⋅‖L2(R)‖x‖). Thus we have shown that ‖M(z)‖≤‖T‖.

Best wishes, Hendrik

From an old version of this post, just kept for documentation:

I just indicate how to prove the boundedness of M. One shows that the multiplicator Ma:=M(⋅+ia) corresponds to the causal autonomous operator Ta:=e−iamTeiam, for all a∈R; note that ‖Ta‖=‖T‖. Similarly, the multiplicator Mr:=M(r⋅) corresponds to the causal autonomous operator ˜Tr:=S1/rTSr for all r>0, where Srf:=f(r⋅); note that ‖˜Tr‖=‖T‖. Using these two transformations one shows that ‖M(z)‖=‖˜M(1)‖, with some ˜M corresponding to a rescaled operator ˜T that has the same norm as T. Finally, (1) implies that ‖˜M(1)‖≤c‖˜T‖=c‖T‖ with a constant c depending neither on M nor on z.

Dear virtual lecturers,

concerning the proof of Lemma 8.1.1: For the second part (starting on p.95) I prefer less computational arguments. On (−∞,0) the inequality supν>0∫(−∞,0)‖f(t)‖2e−2νtdt<∞ together with the monotone convergence theorem (Beppo Levi) implies that g(t):=limν→∞‖f(t)‖2e−2νt (t<0) defines a function g∈L1(−∞,0). Then [g=∞] is a null set, and on the complement of this null set one has f(t)=0. And on [0,∞) the inequality supν>0∫[0,∞)‖f(t)‖2e−2νtdt<∞ together with the monotone convergence theorem (Beppo Levi) implies that limν→0‖f(t)‖2e−2νt=‖f(t)‖2 yields a function ‖f(⋅)‖2∈L1([0,∞)).

Concerning Theorem 8.1.2: As a preparation and motivation I would have expected a statement that, for any f∈L2([0,∞);H), the definition g(i⋅+ν):=Lνf(ν>0) yields an element of H2(CRe>0;H), with the equality of the norms on the last line of Theorem 8.1.2. (The holomorphy of g follows from the formula g(z)=1√2π∫e−izxf(x)dx (Rez>0).) These properties, together with the linearity of the mapping f↦g thus defined also yields the asserted uniqueness statement in the theorem (whose proof I didn't see in the lecture notes).

(I hope I didn't overlook some part of the lecture notes where some of these properties had already been mentioned. But in this case a hint to these parts should have been included.)

Best wishes, Jürgen