In the previous section, we studied the convergence of measures such as Iθ, µθ, gθ, Pθ, and Rθ and more generally (PED)-measures, including hθ, with θ fixed and admissible (in the sense that the corresponding measure is well-defined). Taking now θ variable leads to pointwise convergence of bundles.
We recall the following definition, adapted to our needs, from (Egghe & Rousseau,
2022).
A bundle m is a set of functions, referred to as measures, mθ, with θ belonging to a subset of [0, +∞], detailed further on. These measures are defined on a subset Z ⊂ U. For fixed Z ∈ Z we have a function θ → mθ(Z), where now θ ranges in a subset of [0, +∞], depending on Z.
When studying two functions Z and Y at the same time, we will always assume that θ belongs to the set where mθ(Z), as well as mθ(Y), are defined. We simply write “all admissible θ”. We refer the reader to the previous sections for examples of measures mθ.
Without new proofs necessary, we have the following theorem.
If (Zn)n → Z, pointwise, with all Zn in U, then
(a) (I(Zn))n → I(Z), pointwise, with, for Y in U, I(Y): θ→ Iθ(Y).
(b) (µ(Zn))n → µ(Z), pointwise, with, for Y in U, µ(Y): θ→ µθ(Y).
(c) (g(Zn))n → g(Z), pointwise, with, for Y in U, g(Y): θ→ gθ(Y).
(d) (h(Zn))n → h(Z), pointwise, with, for Y in U, h(Y): θ→ hθ(Y).
(e) (P(Zn))n → P(Z), pointwise, with, for Y in U, P(Y): θ→ Pθ(Y) = Y(θ).
(f) (R(Zn))n → R(Z), pointwise, with, for Y in U, R(Y): θ→ Rθ(Y).
(g) A similar result holds for (PED)-bundles.
We next show how parts (a), (b), and (c) can be strengthened. In this proof, we say that θ is admissible if gθ(Z) and all gθ(Zn) exist.
If (Zn)n → Z, pointwise, with all Zn in U, then
(a) (I(Zn))n → I(Z), uniformly in θ.
(b) (µ(Zn))n → µ(Z), uniformly in θ.
(c) if Z is continuous and Z≠0, then (g(Zn))n → g(Z), uniformly in θ, with θ admissible.
(d) Point (c) is not valid if Z = 0.
(a) As |Zn – Z| tends to zero and all functions Zn and Z are decreasing, each function Zn-Z is integrable. The functions Zn-Z are moreover uniformly bounded, hence it follows from Theorem 2 (based on Arzelà's theorem) that $ \int_{0}^{T}\left|Z_{n}(s)-Z(s)\right| ds \rightarrow 0 $.
As $ \left|\int_{0}^{\theta}\left(Z_{n}(s)-Z(s)\right) ds\right| \leq \int_{0}^{\theta}\left|Z_{n}(s)-Z(s)\right| ds \leq \int_{0}^{T}\left|Z_{n}(s)-Z(s)\right| ds $ this proves that $\int_{0}^{\theta} Z_{n}(s) ds \rightarrow \int_{0}^{\theta} Z(s) ds$ uniformly in θ.
(b) For n fixed, µ(Zn) is decreasing in the variable θ (because all Zn are decreasing). Moreover, it follows from point (a) that µ(Zn) is pointwise decreasing to µ(Z) in the variable θ. Now µ(Z) is differentiable, hence continuous in θ. Applying now the second Dini theorem on the sequence fn = µ(Zn) proves that (µ(Zn))n → µ(Z), uniformly in θ.
(c) Put x = gθ(Z) then we know, see the proof of Theorem 3, that
$ \left|g_{\theta}\left(Z_{n}\right)-g_{\theta}(Z)\right| \leq \frac{1}{\theta x}\left|\int_{0}^{x} Z_{n}(s) ds-\int_{0}^{x} Z(s) d s\right| $. As Z is decreasing, we know that $ \theta x=\frac{1}{x} \int_{0}^{x} Z(s) ds \geq \mu_{T}(Z) $, hence: $\left|g_{\theta}\left(Z_{n}\right)-g_{\theta}(Z)\right|=\frac{1}{\mu_{T}(Z)}\left|\int_{0}^{x} Z_{n}(s) ds-\int_{0}^{x} Z(s) ds\right|$. Using part (1) of this theorem this proves part (2).
(d). Part (c) is not valid for Z = 0. We provide a counterexample. Let Zn be the constant function an on [0,T] with $ \lim_{n \rightarrow \infty} a_{n} $ = 0. Then (Zn)n tends uniformly, hence pointwise, on [0,T] to the function Z = 0. As, gθ(Zn) = hθ(Zn) = an/θ, we see that this sequence tends to gθ(Z) = hθ(Z)=0 pointwise, but not uniformly.
We next investigate what happens if (Zn)n → Z, uniformly, with again all Zn, and hence Z, in U.
If (Zn)n → Z, uniformly, with all Zn in U, then
(a) (I(Zn))n → I(Z), uniformly in θ.
(b) (µ(Zn))n → µ(Z), uniformly in θ.
(c) If Z ≠ 0, then also (g(Zn))n → g(Z), uniformly.
(d) If θ0 = inf{θ; θ is admissible} > 0, then (h(Zn))n → h(Z), uniformly. This property does not hold if θ0 = 0 (this happens e.g. when Z(T)=0).
(e) (P(Zn))n → P(Z), uniformly.
(f) If θ0 =inf{θ; θ is admissible} > 0 then (R(Zn))n → R(Z), uniformly. This property does not hold if θ0 = 0 (this happens e.g. if Z(T)=0).
Points (a), (b), and (c) are just special cases of theorem 8 because uniform convergence implies pointwise convergence.
(d). From the proof of Theorem 4, with fm(s) = θs, x = hθ(Z) and xn = hθ(Zn)we know:
$ \theta\left|x-x_{n}\right| \leq\left|Z(x)-Z_{n}(x)\right|. $
Hence: $ \left|h_{\theta}(Z)-h_{\theta}\left(Z_{n}\right)\right| \leq \frac{1}{\theta}\left|Z(x)-Z_{n}(x)\right| $
This shows that the uniform convergence of (h(Zn))n follows from the uniform convergence of (Zn)n and the fact that θ0 > 0.
(e). This is trivial as $ \left|P_{\theta}\left(Z_{n}\right)-P_{\theta}(Z)\right|=\left|Z_{n}(\theta)-Z(\theta)\right| $
(f). From Theorem 6 we know that $ \left|R_{\theta}^{2}\left(Z_{n}\right)-R_{\theta}^{2}(Z)\right| $
$ \leq\left|\int_{0}^{h_{\theta}(Z)} Z_{n}(s) ds-\int_{0}^{h_{\theta}(Z)} Z(s) ds\right|+\left|\int_{h_{\theta}(Z)}^{h_{\theta}\left(Z_{n}\right)} Z_{n}(s) ds\right| $
Now, when n increases, the first term in this sum converges uniformly to zero because of the part (a) – using hθ(Z) as θ – while the second term is smaller than $ \max_{n}\left(Z_{n}(0), Z(0)\right) *\left|h_{\theta}\left(Z_{n}\right)-h_{\theta}(Z)\right| $. This term converges uniformly to zero because of point (d).