Proof explanation: Prove that $\Vert f(b)-f(a)-f'(a)(b-a) \Vert\leq \sup_{x\in [a,b]} \Vert f''(x)\Vert\Vert b-a\Vert^2$

Question

Two days ago, I asked a question Prove that $\Vert f(b)-f(a)-f'(a)(b-a) \Vert\leq \sup_{x\in [a,b]} \Vert f''(x)\Vert\Vert b-a\Vert^2$ 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.

Answer 1

Recall Taylor's formula for $g\colon[0,1]\rightarrow\mathbb{R}^m$ twice differentiable $$ g(1)=g(0)+g'(0)+\int_0^1g^{\prime\prime}(s)(1-s)\mathrm{d}s; $$ (I will prove this ad hoc bellow). Now apply this to the function $g(t):= f((1-t)a+tb)$ for $t\in[0,1]$ to get your answer:

Answer: By the chain rule: $$ f(b)=f(a)+f^\prime(a)(b-a)+\int_0^1(1-t)\langle a-b,f^{\prime\prime}((1-t)a+tb)(a-b)\rangle\mathrm{d}t $$ so that by the triangle inequality for integrals $$ \|f(b)-f(a)-f^\prime(a)(b-a)\|\leq \int_0^1 (1-t)\|f^{\prime\prime}((1-t)a+tb)(b-a)\|\|b-a\|\mathrm{d} t\\ \leq \sup_{x\in[a,b]}\|f^{\prime\prime}(x)(b-a)\|\|b-a\|\int_0^1 (1-t)\mathrm{d}t\\ =\frac{1}{2}\sup_{x\in[a,b]}\|f^{\prime\prime}(x)(b-a)\|\|b-a\| $$ where in the first inequality we applied Cauchy-Schwarz. Now it seems to me that one can get $\frac{1}{2}$ 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 $\mathbb{R}^{n\times n}$ (i.e., where $f^{\prime\prime}$ sits). However, let's say that we work with Euclidean (square) norms only. Let $M\in\mathbb{R}^{n\times n}$ and $x\in\mathbb{R^n}$. Then, again by Cauchy-Schwarz, $$ \|Mx\|^2=\sum_{i=1}^n (ROW_jM\cdot x)^2\leq \sum_{i=1}^n \|ROW_jM\|^2\| x\|^2=\|M\|^2\|x\|^2, $$ so that we can substitute $M=f^{\prime\prime}(a)$, $x=b-a$ to get $$ \|f(b)-f(a)-f^\prime(a)(b-a)\|\leq\frac{1}{2}\sup_{x\in[a,b]}\|f^{\prime\prime}(x)\|\|b-a\|^2. $$ Proof of Taylor's formula: Let $$ I=\int_0^1\dfrac{\mathrm{d}}{\mathrm{d}s}[(1-s)g^\prime(s)]\mathrm{d}s, $$ which we compute in two ways. First, by FTC $$ I=-g^\prime(0). $$ Second, by the product rule and linearity of integration, $$ I=\int_0^1-g^\prime(s)\mathrm{d}s+\int_0^1(1-s)g^{\prime\prime}(s)\mathrm{d}s=-g(1)+g(0)+\int_0^1(1-s)g^{\prime\prime}(s)\mathrm{d}s. $$ 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 $\mathbb{R}^n$ and $\mathbb{R}^m$).

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 $o(|b-a|^2)$.