Convergence from some recursive hierarchy inequality (II)

Slightly simplifying the problem

$t=0$ implies $f^N_n(t)=0$ for any n<N :

We can prove that for any $n<N$, $f^N_n(t)=0+0$ because it's the sum of an empty integral and something multiplied by $t=0$.

We don't need $T$ :

Let's assume that the $f^N_n$ are satisfying the inequality in the interval $[0,T]$.

Then we can define functions $g^N_n$ by :

• $g^N_n(t)=f^N_n(t), \quad\forall t \in [0,T]$ and
• $g^N_n(t)=f^N_n(T), \quad\forall t \in [T,+\infty[$.

Then, let $t\in [T,+\infty[, n<N\in ℕ$. $g(t)=f(T)=g(T)\le Cn\int_0^T \big(g_{n+1}^N(u)+g_n^N(u)\big)du + \frac{Cn^2}{\sqrt{N-n}}T$ $\le Cn\int_0^t \big(g_{n+1}^N(u)+g_n^N(u)\big)du + \frac{Cn^2}{\sqrt{N-n}}t$ because all the terms are positive.

Then, for simplicity, we can assume that the $f_n^N$ are from $\mathbb{R}^+$ to $[0,2]$.

Upper bound focusing on the integrals

Here, I write $M$ instead of $N$ on purpose, because I need both for next part.
Let's define some other functions $F^M_n:\mathbb{R}^+\to[0,2]$ similar to $f$ :

• $F^M_M(t)=2, \quad\forall t \in \mathbb{R}^+$
• $F^M_n(t)=\min\big(2,\quad Cn\int_0^t F^M_n(u)+F^M_{n+1}(u)du\big), \quad\forall T \in \mathbb{R}^+, \quad n\le M-1$.

Let's now create another set of functions $G^M_n:\mathbb{R}^+\to[0,2]$. The $G^M_n$ are designed to be upper bounds to the $F^M_n$ (not directly to the $f^M_n$, but we will care about it later), and with more easily computable values (we use multiplications instead of recursive integrals). For $M\in\mathbb{N}$ :

• $T^M_M=0, \quad T^M_{M-1}=\frac{1}{4C(M-1)}$.
• $T^M_n=T^M_{n+1}+\frac{1}{8Cn}, \quad\forall n\le M-2$.
• Note that $(T^M_n)_n$ is strictly decreasing.
• $G^M_M(t)=2, \quad\forall t \in \mathbb{R}^+$.
• $G^M_{M-1}(t)=4C(M-1)t, \quad\forall t \in [0,2T^M_{M-1}]$.
• $G^M_{M-1}(t)=2, \quad\forall t \in [2T^M_{M-1},+\infty[$.
• $G^M_n(t)=\frac{1}{2}G^M_{n+1}(t), \quad\forall t \in [0,T^M_{n+1}], \quad n\le M-2$.
• $G^M_n(t)=\frac{1}{2}+4Cn(t-T^M_{n+1}), \quad\forall t \in [T^M_{n+1}, T^M_{n+1}+3(T^M_n-T^M_{n+1})], \quad n\le M-2$.
• $G^M_n(t)=2, \quad\forall t \in [T^M_{n+1}+3(T^M_n-T^M_{n+1}),+\infty[, \quad n\le M-2$.

Now, let's observe a few properties functions $G^M_n$ have :

• $G^M_n(T^M_M)=G^M_n(0)=0, \quad\forall n\le M-1$.
• $G^M_{M-1}(T^M_{M-1})=\frac{4C(M-1)}{4C(M-1)}=1$.
• $G^M_n(T^M_n)=G^M_n(T^M_{n+1}+(T^M_n-T^M_{n+1}))=\frac{1}{2}+\frac{4Cn}{8Cn}=1, \quad\forall n\le M-2$.
• By induction, $G^M_n(T^M_m)=\frac{1}{2^{m-n}}, \quad\forall n\le m\le M-1$.
• For $n\le M-1$, on each interval $[T^M_{m+1}, T^M_m], m\le n$ and $[T^M_n, T^M_n+2(T^M_n-T^M_{n+1})]$ and $[T^M_n+2(T^M_n-T^M_{n+1}),+\infty[$, $G^M_n$ is an affine function. In other words, $G^M_n$ is piecewise affine.
• For any $n$, $t\mapsto G^M_n(t)$ is a continuous non-decreasing function.

Proposition 1 :
$G^M_n(t)\ge \min\big(2,\quad Cn\int_0^t G^M_n(u)+G^M_{n+1}(u)du\big), \quad\forall t \in \mathbb{R}^+, \quad n\le M-1$.
When $G^M_n(t)=2$, the result is directly true. Let's focus on the case where $G^M_n(t)<2$, i.e. $t\in [0,T^M_n+2(T^M_n-T^M_{n+1})[$.

Proof for $n=M-1$ :
Let $t\in [0,T^M_{M-1}]$, then $G^M_{M-1}(t)=4Cnt=Cn\int_0^t 2+2du \ge Cn\int_0^t G^M_M(u)+G^M_{M-1}(u)du$ because we always have $G^M_n(t)\le 2$. $\quad\square$

Proof for $n\le M-2$ :
We well prove it on intervals of the form $[T^M_{M-k}, T^M_{M-k-1}]$, by induction on $k$.

Initialization :
When $k=0$, we mimic the proof given for $n=M-1$ : Let $t\in [T^M_M,T^M_{M-1}]=[0,T^M_{M-1}]$.
$G^M_n(t)=\frac{4Cnt}{2^{(M-1)-n}} =Cn\int_0^t\frac{1}{2^{(M-1)-(n+1)}}+\frac{1}{2^{(M-1)-(n+1)}}du$ $\ge Cn\int_0^t\frac{1}{2^{(M-1)-(n+1)}}+\frac{1}{2^{(M-1)-n}}du = Cn\int_0^t G^M_{n+1}(T^M_{M-1})+G^M_n(T^M_{M-1})du\quad(*)$ $\ge Cn\int_0^t G^M_{n+1}(u)+G^M_n(u)du$, because the $G^M_m$ are non-decreasing.

$(*)$ This inequality is actually strict when $t>0$.
We will need this point later, on the proof of proposition 2.

Heredity :
Let's assume that, for some $k\le M-n-2$, we have $G^M_n(T^M_{M-k})\ge Cn\int_0^{T^M_{M-k}} G^M_n(u)+G^M_{n+1}(u)du$. Let $t\in [T^M_{M-k}, T^M_{M-(k+1)}]$. $G^M_n(t)=G^M_n(T^M_{M-k})+\frac{t-T^M_{M-k}}{T^M_{M-(k+1)}-T^M_{M-k}}\times \big(G^M_n(T^M_{M-(k+1)})-G^M_n(T^M_{M-k})\big)$ $=G^M_n(T^M_{M-k})+\frac{t-T^M_{M-k}}{1/(8C(M-(k+1)))}G^M_n(T^M_{M-k})$ $\ge G^M_n(T^M_{M-k})+(t-T^M_{M-k})(8Cn)G^M_n(T^M_{M-k})$ $\ge G^M_n(T^M_{M-k})+(t-T^M_{M-k})Cn(G^M_n(T^M_{M-(k+1)})+G^M_{n+1}(T^M_{M-(k+1)}))$ $\ge Cn\int_0^{T^M_{M-k}} G^M_n(u)+G^M_{n+1}(u)du \quad + \quad Cn\int_{T^M_{M-k}}^t G^M_n(u)+G^M_{n+1}(u)du$.
The induction hypothesis is used in the last inequality. $\quad\square$

Heredity when $k=M-n-1$:
Let $t\in [T^M_{n+1},T^M_n+2(T^M_n-T^M_{n+1})]$. $G^M_n(t)=\frac{1}{2}+4Cn(t-T^M_{n+1}) =G^M_n(T^M_{n+1})+Cn\int_{T^M_{n+1}}^t 2+2 du$ $\ge Cn\int_0^{T^M_{M-k}} G^M_n(u)+G^M_{n+1}(u)du \quad + \quad Cn\int_{T^M_{M-k}}^t G^M_n(u)+G^M_{n+1}(u)du$.
The induction hypothesis is used (again) in the last inequality. $\quad\square$

Once the behavior of $G$ is understood, we can easily compare it to $F$.

Proposition 2 :
$F^M_n(t)\le G^M_n(t), \quad\forall M, n, t$.
The equality holds only when $t=0$ or $M=n$ or $G^M_n(t)=2$.

Proof :
By contradiction, let $n$ be the biggest number the proposition is wrong. By $(*)$, we know that there is $\epsilon>0$ such that $\forall t\in ]0,\epsilon], G^M_n(t)>F^M_n(t)$.

Let $t>0$ be the smallest number where $F^M_n(t)=G^M_n(t)<2$. Actually, $t$ should be the "inf" not the "min", but here it's actually the same because both $F^M_n$ and $G^M_n$ are continuous.

By proposition 1, $G^M_n(t)\ge Cn\int_0^t G^M_n(u)+G^M_{n+1}(u)du$
$> Cn\int_0^t F^M_n(u)+F^M_{n+1}(u)du = F^M_n(t)$ (the last inequality is because $t$ is minimal).
That's a contradiction. $\quad\square$

Now, we can prove that the limit is zero. Not already for $f$ but at least for $F$, using an harmonic serie.

Proposition 3 :
$lim_{M\to +\infty}F^M_n(t)=lim_{M\to +\infty}G^M_n(t)=0, \quad\forall t\in\mathbb{R}^+,\quad n\in\mathbb{N}$.

Proof
Let $t\in\mathbb{R}^+, n\in\mathbb{N}, C>0$. As long as $0\le F^M_n(t)\le G^M_n(t)$, we only need to prove the result for $G$.

Let's find an estimation of $T^M_n$.
$T^M_n=T^M_{M-1}+\sum_{i=n}^{M-2}(T^M_i-T^M_{i-1}) =\frac{1}{4C(M-1)}+\sum_{i=n}^{M-2}\frac{1}{8Cn} =\frac{\log(M)-\log(n)+O(1)}{4C}$.
In particular, it implies that $\lim_{M\to +\infty}T^M_n=+\infty$.

We can take $T^M_n$ big enough to be greater than $t$. $G^M_n(T^M_n)=1$ by construction, and $G^M_n$ is convex on $[0,T^M_n]$ ($G^M_n$ is differentiable almost everywhere, and $G^{M'}_n$ is non-decreasing where it's defined). So, $G^M_n(T^M_n)\le\frac{t}{T^M_n}\to 0$ because $T^M_n\to+\infty$. $\quad\square$

Upper bound for the second term

$f^N_n$ is defined almost the same way $F^M_n$ is, except that it has a second term $\frac{Cn^2}{\sqrt{N-n}}t$. When $n<<N$, those terms are very small.

Let's assume $M<<N$. In particular, we can suppose that $2M<N$. Then $\sqrt{N-n}>\sqrt{N/2}, \quad\forall n\le M$.

The problem can be solved by using one more set of functions defined when $N>2M$ :

• $H^M_M(t,N)=0, \quad\forall t,\quad 2M<N$.
• $H^M_n(t,N)=Cn\int_0^t H^M_n(u,N)+H^M_{n+1}(u,N)du +\frac{Cn^2}{\sqrt{N/2}}t, \quad\forall n\le M-1$.

Then, when $n, M, t$ are fixed, $H^M_n(t,N)=O(\frac{1}{\sqrt{N}})$, because only linear operations (sum, integral, multiplication by something independent on $N$) are applied on $\frac{1}{\sqrt{N/2}}$.

In other words, $H^M_n(t,N)\to 0$ when $N\to +\infty$.

Lemma 4 :
Let $a, b, c\in\mathbb{R}^+$, then $\min(a,b)+c\ge\min(a,b+c)$.

Proof :

• If $a\le b$, then $\min(a,b)+c=a+c\ge a\ge\min(a,b+c)$.
• If $b\le a$, then $\min(a,b)+c=b+c\ge\min(a,b+c)$. $\quad\square$

Proposition 5 :
Let $n,M,N\in\mathbb{N}$ with $n<M<N/2$, and $t\in\mathbb{R}^+$.
$G^M_n(t)+H^M_n(t,N)$
$\ge \min(2, Cn\int_0^t G^M_n(u)+H^M_n(u,N)+G^M_{n+1}(u)+H^M_{n+1}(u,N)du+\frac{Cn^2}{\sqrt{N/2}}t)$.

Proof :
$G^M_n(t)+H^M_n(t,N)$
$\ge \min\big(2, Cn\int_0^t G^M_n(u)+G^M_{n+1}(u)du\big)+Cn\int_0^t H^M_n(u, N)+H^M_{n+1}(u,N)du+\frac{Cn^2}{\sqrt{N/2}}t$
$\ge \min(2, Cn\int_0^t G^M_n(u)+H^M_n(u,N)+G^M_{n+1}(u)+H^M_{n+1}(u,N)du+\frac{Cn^2}{\sqrt{N/2}}t)$
The first inequality comes from proposition 1 and the definition of H. The second inequality comes from lemma 4. $\quad\square$

By a reasoning similar to proof of proposition 2, we obtain that :

Proposition 6 :
$0\le f^N_n(t)\le G^M_n(t)+H^M_n(t,N)$ when $n<M<N/2$, and the equality holds only if $t=0$ or $f^N_n(t)=2$.

Proof
Let $M, N\in\mathbb{N}$ with $M<N/2$.
By contradiction, let's assume that proposition 6 is false for some $t>0$ and $n<M$. Since n is a natural number with an upper bound, we can take the greatest $n$ (with $M, N$ fixed) where proposition 6 is false.

To simplify the notations, I will write $GH^M_n(t,N)$ instead of $G^M_n(t)+H^M_n(t,N)$.
Let's also take $t>0$ be the first time $f^N_n(t)=GH^M_n(t,N)$. As long as it's a proof by contradiction, we also need to assume that $f^N_n(t)<2$.

We need to make sure $t$ is defined. Since $\frac{Cn^2}{\sqrt{N/2}}>\frac{Cn^2}{\sqrt{N-n}}$ and $f^N_{n+1}(0)=f^N_{n+1}(0)=0$ and the $f$ are continuous, there exists $\epsilon>0$ such that
$\forall u\in ]0,\epsilon],\quad Cn(f^N_n(u)+f^N_{n+1}(u))<\frac{Cn^2}{\sqrt{N/2}}-\frac{Cn^2}{\sqrt{N-n}}$.
It means $\forall u\in]0,\epsilon],\quad f^N_n(u)<H^M_n(u,N)\le GH^M_n(u,N)$.
So, $E:=\{u>0|f^N_n(u)=GH^M_n(u,N)\}\subset ]\epsilon,+\infty[$. Now we can take $t=\inf(E)=\min(E)$ because $f^N_n$ and $GH^M_n$ are continuous.
(In other words, we supposed by contradiction that $E\neq\emptyset$, and $f^N_n(t)<2$)

By the choice of $n$ and $t$ we have :
$f^N_n(t)\le Cn\int_0^t f^N_n(u)+f^N_{n+1}(u)du+\frac{Cn^2}{\sqrt{N-n}}t$, by the definition of $f$
$\le Cn\int_0^t f^N_n(u)+f^N_{n+1}(u)du+\frac{Cn^2}{\sqrt{N/2}}t$, because $n<M<N/2$
$< Cn\int_0^t GH^M_n(u,N)+GH^M_{n+1}(u,N)du+\frac{Cn^2}{\sqrt{N/2}}t$, because $t$ is minimal
$\le GH^M_n(t,N)$, this comes from proposition 5 (here I didn't wrote the $\min(2,\cdots)$ part because we assumed $GH^M_n(t,N)=f^N_n(t)<2$).

So, we have both $f^N_n(t)=GH^M_n(t,N)$ (definition of $t$) and $f^N_n(t)<GH^M_n(t,N)$ : contradiction. $\quad\square$

Knowing that both $G^M_n(t)\to 0$ when $M\to +\infty$ and $H^M_n(t,N)\to 0$ when $M$ is fixed and $N\to +\infty$, we conclude that $f^N_n(t)\le G^M_n(t)+H^M_n(t,N)\to 0$ when $N\to +\infty$. $\quad\square$