# Alternative proof of Taylor's formula by only using the linear approximation property

So a function $$f: E \to F$$ between the normed spaces $$E,F$$ is called differentiable in $$x \in E$$ if there exists a bounded linear map $$Df(x): E \to F$$ such that for every $$h \in E$$ we have $$f(x+h)=f(x)+Df(x)h + o(||h||). \tag{1}$$ If $$f$$ is differentiable for every $$x \in E$$ and $$Df: x \mapsto Df(x)$$ is differentiable for every $$x \in E$$ too we get analogously $$Df(x+e)=Df(x)+D^2f(x)e+o(||e||). \tag{2}$$ Then $$f$$ is called twice differentiable and for every $$h\in E$$ we have the "Taylor expansion of second degree" $$f(x+h)=f(x)+Df(x)h+\frac{1}{2}D^2f(x)[h] + o(||h||^2), \tag{3}$$ where $$D^2f(x)[h]:=(D^2f(x)h)h$$ for better readability.

I have two questions:

• How can $$(3)$$ be proven without resorting to the "standard proof" of using integrals? I want to show it by only using the linear approximations given in $$(1)$$ and $$(2)$$. Inserting $$(2)$$ in $$(1)$$ doesn't result in something useful though. Can this be done?
• Can $$(3)$$ be used as an alternative definition off twice-differentiability? Analogously what about the general case of $$n$$-times differentiability: $$f(x+h) = f(x) + \sum_{j=1}^{n} \frac{1}{j!} D^jf(x)[h] + o(\|h\|^n) \tag{4}$$
• If integrals are forbidden, is it fair game to use the mean value inequality ? Jun 26 '19 at 19:57
• I'd say that in this case, using Landau notation makes things more difficult, and not less. As you don't have a precise formula for $o(||h||)$, you can't manipulate (1) and (2) to let the error terms cancel. Jun 28 '19 at 5:48
• @GabrielRomon Sorry, I didn't see your comment. Just using the non-integral version of the mean-value inequality would be fine. Jun 28 '19 at 21:30

A few days back I wrote an answer with some detail about Taylor polynomials for maps between Banach spaces. You can see my answer here. What I proved is that (I'm sorry about the differences in notation)

Taylor Expansion Theorem:

Let $$V$$ and $$W$$ be Banach spaces over the field $$\Bbb{R}$$, let $$U$$ be an open subset of $$V$$, and fix a point $$a \in U$$. Let $$f:U \to W$$ be a given function which is $$n$$ times differentiable at $$a$$ (in the Frechet differentiable sense). Define the Taylor polynomial $$T_{n,f}:V \to W$$ by $$\begin{equation} T_{n,f}(h) = f(a) + \dfrac{df_a(h)}{1!} + \dfrac{d^2f_a(h)^2}{2!} + \dots + \dfrac{d^nf_a(h)^n}{n!} \end{equation}$$ Then, $$f(a+h) - T_{n,f}(h) = o(\lVert h \rVert^n)$$.

Explicitly, the claim is that for every $$\varepsilon > 0$$, there is a $$\delta > 0$$ such that for all $$h \in V$$, if $$\lVert h \rVert < \delta$$, then $$\begin{equation} \lVert f(a+h) - T_{n,f}(h) \rVert \leq \varepsilon \lVert h \rVert^{n}. \end{equation}$$

The proof is pretty short if you know what you're doing. The idea is to use induction, and most importantly, the mean-value inequality for maps between Banach spaces. I don't think it is possible to derive $$(3)$$ directly from $$(1)$$ and $$(2)$$ alone, because $$(2)$$ talks about how the derivative $$Df$$ changes, while $$(1)$$ talks about how the function $$f$$ changes, and ultimately $$(3)$$ talks about how much $$f$$ changes. So, you somehow have to relate changes in $$Df$$ to changes in $$f$$... this is roughly speaking, what the mean-value inequality does.

The proof I showed in my other answer is pretty much from Henri Cartan's excellent book Differential Calculus. Also, Henri Cartan's book has a proof of the mean-value inequality which doesn't rely on integrals. Alternatively, you can take a look at Loomis and Sternberg's book Advanced Calculus. Here, they prove the mean value inequality in a rather elementary way without integrals, and it's also a relatively short proof. It is proven in Chapter 3, Theorem 7.4 (which uses theorem 7.3); this is on page 148-149 of the book (I prefer this proof to Cartan's proof).

For your other question, I assume you mean the following:

Does the existence of a polynomial $$P$$, which equals $$f$$ up to order $$2$$ at $$x$$ imply that $$f$$ is twice differentiable at $$x$$? Or more precisely, does the existence of a continuous linear map $$A_1:E \to F$$, and a symmetric continuous bilinear map $$A_2:E \times E \to F$$ such that for all $$h \in E$$, $$\begin{equation} f(x+h) = f(x) + A_1(h) + A_2(h,h) + o(\lVert h \rVert^2) \end{equation}$$ imply that $$f$$ is twice Frechet differentiable at $$x$$?

The answer to this question is no. We can see this even in the single variable case (this following example is from Spivak's Calculus, page 413, 3rd edition). Take $$E=F=\Bbb{R}$$, and define $$f: \Bbb{R} \to \Bbb{R}$$ by $$\begin{equation} f(x) = \begin{cases} x^{n+1}& \text{if x irrational} \\ 0 & \text{if x rational} \end{cases} \end{equation}$$ ($$n\geq 2$$). Then choose $$x=0$$ and the zero polynomial $$P \equiv 0$$. It is easy to verify that $$\begin{equation} f(0 + h) = 0 + o(|h|^n) \end{equation}$$ However, if $$a \neq 0$$, then $$f'(a)$$ doesn't exist so $$f''(0)$$ is not even defined. Hence, what this shows is that the existence of a well-approximating polynomial does not guarantee that the function is sufficiently differentable.

• What if $f$ is assumed to be continuous? Can there still be a second-order polynomial approximating $f$ without it being twice-differentiable? Jun 29 '19 at 11:44
• @JannikPitt There is a "partial converse" to Taylor's theorem, if you assume enough regularity. For the one-dimensional case,refer to this post: mathoverflow.net/questions/88501/converse-of-taylors-theorem. For the Banach space version, refer to either Abraham Robbin: Transversal mappings and flows (Theorem $2.1$) or see Abraham Marsden Ratiu: Manifolds, Tensor Analysis and Applications, supplement 2.4B (3rd edition). While the statement of the theorem is very nice, the proof seems pretty involved. If you want, I could just add the statement of the converse into my answer. Jun 29 '19 at 12:10

The differential Df(x) is a linear map that approximates your function f, according to your relation (1). And that linear map depends on x (the point around which you consider the Taylor expansion). The map Df , which associates a vector to a linear map doesn't have to be linear indeed. The second differential DDf(x) is a linear map that associates a vector to a linear map , and so on. Your relation (1) tells us that the differential of f exists (definition). Your relation (2) tells us that the differential of Df exists (definition). Your relation (3) tells us that there is a valid finite second order (degree) Taylor expansion of f at x (also a definition). From (1) and (2) you cannot derive (3). That's not always possible even for real function , even more so in this abstract setting. In a valid, finite Taylor expansion, the remainder must grow slower than the last term of the Taylor expansion. And no, (3) is not equivalent to second order differentiability of f.

• Careful: The object $Df(x)$ is a linear map, while $x \mapsto Df(x)$ is a function taking vectors to linear maps, which in general is not linear. So $Df(x+y)≠Df(x)+Df(y)$ (why would the be equal?). Jun 28 '19 at 10:56
• I'll edit my answer , in order to clarify. Jun 28 '19 at 11:30
• The second derivative is a bilinear map (technically it isn't but it can be viewed as such), how should that approximate $f-Df(x)$? Jun 28 '19 at 21:31
• I edited my answer @JannikPitt . Have a look. So from (1) and (2) you cannot derive (3), and (3) is not equivalent to twicw-differentiability. Relation (3), by definition tells us that f has a valid second order Taylor expansion , that's all. It's a definition. It took me a little bit of time, I'm a bit rusty. Jun 28 '19 at 21:57
• Your confusion might be caused by the fact that the existence of a first order Taylor expansion is very similar to the definition of differentiability. That connection is broken for higher orders. Jun 28 '19 at 22:04