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I have read a lemma in "Numerical Integration of Stochastic Differential Equations" of G.N.Milstein, which was stated without any proof as follows:

Suppose that for arbitrary natural number $N$ and $k=0,...,N$, we have $$u_{k+1}\leq (1+Ah)u_k+Bh^p,$$ where $h=T/N,A\geq 0, B\geq 0,p\geq 1,u_k\geq 0,k=0,...,N.$ Then $$ u_k\leq e^{AT}u_0+\frac{B}{A}(e^{AT}-1)h^{p-1} $$ (where for $A=0$, we put $(e^{AT}-1)/A $ equal to zero).

The author said that it was a well-known result, and I tried to proof this lemma by induction as well as looked for it on the Internet; however, I failed. Can anyone help me to proof this lemma, or have any website containing the proof? Thank you all!

FShrike
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Tom Howard
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1 Answers1

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$$\begin{align}u_{k+1}&\le(1+AT/N)u_k+BT^pN^{-p}\\&\le(1+AT/N)[(1+AT/N)u_{k-1}+BT^pN^{-p}]+BT^pN^{-p}\\&=(1+AT/N)^2u_{k-1}+BT^pN^{-p}(1+1+AT/N)\\&\le(1+AT/N)^2[(1+AT/N)u_{k-2}+BT^pN^{-p}]+BT^pN^{-p}(2+AT/N)\\&=(1+AT/N)^3u_{k-2}+BT^pN^{-p}(1+(1+AT/N)+(1+AT/N)^2)\\&\le\cdots\\&\le(1+AT/N)^{k+1}u_0+BT^pN^{-p}\cdot\sum_{m=0}^{k}(1+AT/N)^m\\&=(1+AT/N)^{k+1}u_0+BT^pN^{-p}\cdot\frac{(1+AT/N)^{k+1}-1}{AT/N}\\&=(1+AT/N)^{k+1}u_0+\frac{B}{A}\frac{T^{p-1}}{N^{p-1}}\cdot((1+AT/N)^{k+1}-1)\end{align}$$

Whenever $k\le N$. Using the monotonicity of the limit for $e^x$, and swapping $k$ for $k+1$, we get for all $k\le N$: $$\begin{align}u_k&\le(1+AT/N)^Nu_0+\frac{B}{A}h^{p-1}((1+AT/N)^N-1)\\&\le e^{AT}u_0+\frac{B}{A}h^{p-1}(e^{AT}-1)\end{align}$$Holds.

Hope this helps, feel free to query further.

FShrike
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