Vtyjkyuk

Two days ago, I asked a question
Prove that $Vert f(b)-f(a)-f'(a)(b-a) Vertleq sup_xin [a,b] Vert f''(x)VertVert b-aVert^2$ but was answered just once. However, I am finding it hard to understand the proof provided.

Please, I'll need thorough explanation or another proof. As to the proof, I don't understand

1. why triple sum was used and not double.




2. If we differentiate $g(x)=f(b)-f(x)-f'(x)(b-x)$, what do we get?

I am asking because I find it hard to comprehend the proof. Please, can anyone explain these to me? Alternative proofs are welcome.

Thanks.

Recall Taylor's formula for $gcolon[0,1]rightarrowmathbbR^m$ twice differentiable
$$
g(1)=g(0)+g'(0)+int_0^1g^primeprime(s)(1-s)mathrmds;
$$
(I will prove this ad hoc bellow). Now apply this to the function $g(t):= f((1-t)a+tb)$ for $tin[0,1]$ to get your answer:

Answer: By the chain rule:
$$
f(b)=f(a)+f^prime(a)(b-a)+int_0^1(1-t)langle a-b,f^primeprime((1-t)a+tb)(a-b)ranglemathrmdt
$$
so that by the triangle inequality for integrals
$$
|f(b)-f(a)-f^prime(a)(b-a)|leq int_0^1 (1-t)|f^primeprime((1-t)a+tb)(b-a)||b-a|mathrmd t\
leq sup_xin[a,b]|f^primeprime(x)(b-a)||b-a|int_0^1 (1-t)mathrmdt\
=frac12sup_xin[a,b]|f^primeprime(x)(b-a)||b-a|
$$
where in the first inequality we applied Cauchy-Schwarz. Now it seems to me that one can get $frac12$ as the best constant here (clearly optimal if $n=1$), but the answer is, in any case, quite irrelevant as on finite dimensional spaces all norms are equivalent and you do not state clearly what is the norm on $mathbbR^ntimes n$ (i.e., where $f^primeprime$ sits). However, let's say that we work with Euclidean (square) norms only. Let $MinmathbbR^ntimes n$ and $xinmathbbR^n$. Then, again by Cauchy-Schwarz,
$$
|Mx|^2=sum_i=1^n (ROW_jMcdot x)^2leq sum_i=1^n |ROW_jM|^2| x|^2=|M|^2|x|^2,
$$
so that we can substitute $M=f^primeprime(a)$, $x=b-a$ to get
$$
|f(b)-f(a)-f^prime(a)(b-a)|leqfrac12sup_xin[a,b]|f^primeprime(x)||b-a|^2.
$$
Proof of Taylor's formula: Let
$$
I=int_0^1dfracmathrmdmathrmds[(1-s)g^prime(s)]mathrmds,
$$
which we compute in two ways. First, by FTC
$$
I=-g^prime(0).
$$
Second, by the product rule and linearity of integration,
$$
I=int_0^1-g^prime(s)mathrmds+int_0^1(1-s)g^primeprime(s)mathrmds=-g(1)+g(0)+int_0^1(1-s)g^primeprime(s)mathrmds.
$$
Minor appendix: As it stands, my proof tackles the case $m=1$; however, it is completely trivial to extend it (though it is not entirely clear how/if the constant suffers; this of course depends on your choices of norms on $mathbbR^n$ and $mathbbR^m$).

Philosophy: This problem is in any case 1-dimensional; the LHS is the first order Taylor polynomial with step $b-a$, so of course it gives an error $o(|b-a|^2)$.