Taylor’s formula in Banach spaces

Let $U$ be an open subset of a real Banach space $X$.If $f:U\to ℝ$ is differentiable $n+1$ times on $U$,it may be expanded by Taylor’s formula:

$$f(x)=f(a)+\mathrm{D}f(a)\cdot h+\frac{1}{2!}{\mathrm{D}}^{2}f(a)\cdot h^2+\mathrm{\cdots }+\frac{1}{n!}{\mathrm{D}}^{n}f(a)\cdot h^n+R_n(x),$$(1)

with the following expressions for the remainder term $R_n(x)$:

$R_n(x)$$={\displaystyle \frac{1}{n!}}{\mathrm{D}}^{n+1}f(\eta )\cdot {(x-\eta )}^{n}h$Cauchy form of remainder

$R_n(x)$$={\displaystyle \frac{1}{(n+1)!}}{\mathrm{D}}^{n+1}f(\xi )\cdot h^{n+1}$Lagrange form of remainder

$R_n(x)$$={\displaystyle \frac{1}{n!}}{\displaystyle {\int }_0^1}{\mathrm{D}}^{n+1}f(a+th)\cdot {((1-t)h)}^{n}h𝑑t$integral form of remainder

Here $a$ and $x$ must be points of $U$ such that the line segmentbetween $a$ and $x$ lie inside $U$, $h$ is $x-a$,and the points $\xi$ and $\eta$ lie on the same line segment,strictly between $a$ and $x$.

The $k$th Fréchet derivative of $f$ at $a$ is being denoted by${\mathrm{D}}^{k}f(a)$, to be viewed as a multilinear map $X^k\to ℝ$.The $\cdot h^k$ notation means to evaluate a multilinear mapat $(h,\mathrm{\dots },h)$.

1 Remainders for vector-valued functions

If $Y$ is a Banach space, we may also considerTaylor expansions for $f:U\to Y$.Formula (4) takes the same form,but the Cauchy and Lagrange forms of the remainderwill not be exact;they will only be bounds on $R_n(x)$.That is, for $f:U\to Y$,

$\parallel R_n(x)\parallel$$\le {\displaystyle \frac{1}{n!}}\parallel {\mathrm{D}}^{n+1}f(\eta )\cdot {(x-\eta )}^{n}h\parallel$Cauchy form of remainder

$\parallel R_n(x)\parallel$$\le {\displaystyle \frac{1}{(n+1)!}}\parallel {\mathrm{D}}^{n+1}f(\xi )\cdot h^{n+1}\parallel$Lagrange form of remainder

It is not hard to find counterexamples if we attempt to remove the norm signs or changethe inequality to equality in the above formulas.

However, the integral form of the remainder continues to hold for $Y\ne ℝ$,although strictly speaking it only applies if the integrand is integrable.The integral form is also applicable when $X$ and $Y$ are complex Banach spaces.

Mean Value Theorem

The Mean Value Theorem can be obtainedas the special case $n=0$ with the Lagrange form of the remainder:for $f:U\to Y$ differentiable,

$$\parallel f(x)-f(a)\parallel \le \parallel \mathrm{D}f(\xi )\cdot (x-a)\parallel$$(2)

If $Y=ℝ$, then the norm signs may be removed from(2), and the inequality replaced by equality.

Formula (2) also holds under the muchweaker hypothesisthat $f$ only has a directional derivative along the linesegment from $a$ to $x$.

Weaker bounds for the remainder

If $f:U\to Y$ is only differentiable $n$ times at $a$,then we cannot quantify the remainder by the $n+1$th derivative,but it is still truethat

$$R_n(x)=o({\parallel x-a\parallel }^n)\text{ as }x\to a\text{. }$$(3)

Finite-dimensional case

If $X={ℝ}^m$ and $Y=ℝ$,${\mathrm{D}}^k$ has the following expression in terms of coordinates:

$${\mathrm{D}}^{k}f(a)\cdot ({\xi }_1,\mathrm{\dots },{\xi }_k)=\sum _{i_1,\mathrm{\dots },i_k}\frac{{\partial }^{k}f}{\partial x^{i_1}\mathrm{\cdots }\partial x^{i_k}}{\xi }_1^{i_1}\mathrm{\cdots }{\xi }_k^{i_k},$$

where each $i_j$ runs from $1,\mathrm{\dots },m$ in the sum.

If we collect the equal mixed partials(assuming that they are continuous)then

$$\frac{1}{k!}{\mathrm{D}}^{k}f(a)\cdot h^k=\sum _{|J|=k}\frac{1}{J!}\frac{{\partial }^{|J|}f}{\partial x^J}{h}^J,$$

where $J$ is a multi-index of $m$ components, and each component $J_i$ indicateshow many times the derivative with respect to the $i$th coordinate should be taken,and the exponent that the $i$th coordinate of $h$ should be raised toin the monomial $h^J$.The multi-index $J$ runs through all combinationssuch that $J_1+\mathrm{\cdots }+J_m=|J|=k$ in the sum.The notation $J!$ means $J_1!\mathrm{\cdots }{J}_m!$.

All this is more easily assimilated if we remember that${\mathrm{D}}^{k}f(a)\cdot h^k$ is supposed to be a polynomial of degree $k$.Also $|J|!/J!$ is just the multinomial coefficient.

Taylor series

If ${lim}_{n\to \mathrm{\infty }}{R}_n(x)=0$,then we may write

$$f(x)=f(a)+\mathrm{D}f(a)\cdot h+\frac{1}{2!}{\mathrm{D}}^{2}f(a)\cdot h^2+\mathrm{\cdots }$$(4)

as a convergent infinite series. Elegant as such an expansion is,it is not seen very often,for the reason that higher order Fréchet derivatives, especially in infinite-dimensional spaces,are often difficult to calculate.

But a notable exception occurs if a function $f$ is defined by a convergent “power series”

$$f(x)=\sum _{k=0}^{\mathrm{\infty }}{M}_k\cdot {(x-a)}^k$$(5)

where $\{M_k:k=0,1,\mathrm{\dots }\}$ is a family of continuous symmetric multilinear functions $X^k\to Y$.In this case, the series (5) is the Taylor series for $f$ at $a$.