By S. Bandyopadbyay, S. Maiti

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**Additional info for Application of Calculus - Problems and Solutions**

**Example text**

3 (i) that D(S|λ| ) = D(Sλ ) = D(Uβ Sλ Uβ∗ ). 3) = λv βv (Uβ∗ f )(par(v)) = λv βv β¯par(v) f (par(v)) 0 if v ∈ V ◦ , if v = root . 1) λv βv β¯par(v) = |λv |, v ∈ V ◦. We do this in two steps. Step 1. 3) βu = γ, λv βv = |λv |βpar(v) , v ∈ Des(u) \ {u}. 10)) Indeed, since Des(u) \ {u} = ∞ n=1 Chi n+1 n and par(Chi (u)) ⊆ Chi (u), we can deﬁne the wanted system {βv }v∈Des(u) recursively. 2), and then having deﬁned βw for all w ∈ Chi n (u), we deﬁne βv for every v ∈ Chi n+1 (u) by βv = λ−1 v |λv |βpar(v) whenever λv = 0 and by βv = 1 otherwise.

If Sλ is a densely deﬁned weighted shift on a directed tree T with weights λ = {λv }v∈V ◦ , then the following two conditions are equivalent: (i) D(Sλ∗ ) ⊆ D(Sλ ), (ii) Tu ∈ B( 2 (Chi(u))) for all u ∈ V , and sup Tu < ∞. 4) u∈V Proof. (i)⇒(ii) Suppose that D(Sλ∗ ) ⊆ D(Sλ ). 1. Thus, there exists a constant c > 0 such that Sλ f 2 c( f 2 + Sλ∗ f 2 ), f ∈ D(Sλ∗ ). 5) Sλ eu u∈V 2 |f (u)|2 2 |f (u)|2 + c c u∈V λv f (v) , u∈V v∈Chi(u) f ∈ D(Sλ∗ ). 5), we obtain Sλ ev 2 |f (v)|2 u∈V v∈Chi(u) 2 |f (v)|2 + c|f (root)|2 + c u∈V v∈Chi(u) f ∈ D(Sλ∗ ).

6. If T is a Fredholm directed tree, then (i) ind(T ) 1 provided T is rootless, (ii) ind(T ) = 0 provided T is ﬁnite, (iii) ind(T ) −1 provided T has a root and is inﬁnite. 6. FREDHOLM DIRECTED TREES 35 Proof. Without loss of generality, we can assume that T is inﬁnite (otherwise ind(T ) = 0). 5 we are reduced to showing that ind(T ) −1 whenever T has a root and is inﬁnite (note that the trimmed tree may be ﬁnite, however this case has been covered). 5) card(Chi(u)) − 1 ind(T ) = −1 − −1. u∈V≺ Suppose now that T is not leaﬂess (however still inﬁnite and with root).