# Understanding the definition and notation of the second order differential

We're reaching the end of a differential calculus course I'm taking (distance learning), and I'm realizing that I don't fully understand the objects I'm manipulating. In particular, I'm not quite getting the definition and notation of the second order differential: $$[d^2f(a)](h,k):=\big[[d(df)(a)](h)\big](k)$$

I can see (I think, although I'm starting to have doubts) how the first order works $$[df(a)](h)$$: $$df(a)$$ is a linear function that satisfies $$\lim_{h\to0}\frac{f(a+h)-f(a)-[df(a)](h)}{\Vert h\Vert}=0.$$

In the case of a univariate real function, for instance, one could write (using a first order Taylor expansion)
$$[df(a)](h)=f'(a)(h-a)+f(a)$$ a "linear" function of $$h$$, tangential to $$f$$ at $$a$$, whose slope is that of $$f$$ at $$a$$.

Now, trying to decipher the notation of the second order differential: I interpret $$\big[[d(df)(a)](h)\big](k)$$ to mean $$\big[d[df(a)](h)\big](k)$$ in which $$h$$ would be a point by which the linear function $$d[df(a)](h)$$ of $$k$$ is defined. But the notation $$[d^2f(a)](h,k)$$ suggests that $$a$$ only is a point, and that $$[d^2f(a)]$$ is a (bilinear) function of, well, two variables.

These "questions" came as I was trying to understand the following relation, given for the univariate real function case: $$[d^2f(a)](h,k)=hkf''(a)$$ which I really don't. I'm sorry if this a bit of a mess, but so it is in my mind _:). Any help appreciated!

EDIT: Ivo, thank you very much for the level of detail.

• Yes, thanks. I sort of have, but I'm still unable to make it work. I'll edit my post add this as soon as I manage to work it out.
– Mogu
Dec 26 '20 at 10:42

Assume that $$f:\Bbb R^n \to \Bbb R^k$$ is given. For $$p\in \Bbb R^n$$, we have a linear map $$Df(p):\Bbb R^n \to \Bbb R^k$$. This gives rise to a (non-linear!) map $$Df:\Bbb R^n \to {\rm Lin}(\Bbb R^n,\Bbb R^k)$$. This last set is a vector space (isomorphic to $$\Bbb R^{kn}$$), and so we can look at $$D(Df)(p):\Bbb R^n \to {\rm Lin}(\Bbb R^n,\Bbb R^k).$$Given $$v \in \Bbb R^n$$, we have $$D(Df)(p)(v) \in {\rm Lin}(\Bbb R^n,\Bbb R^k)$$, which is to say that $$D(Df)(p)(v)(w) \in \Bbb R^k$$, for $$v,w\in \Bbb R^n$$. This expression is linear in $$v$$ and $$w$$, and since the way of writing it just following the definitions is so horrible, we write it simply as $$D^2f(p)(v,w).$$Then, $$D^2f(p)$$ is a bilinear map taking values in $$\Bbb R^k$$. Why? Because $$D(Df):\Bbb R^n \to {\rm Lin}(\Bbb R^n, {\rm Lin}(\Bbb R^n,\Bbb R^k))$$but $${\rm Lin}(\Bbb R^n, {\rm Lin}(\Bbb R^n,\Bbb R^k))$$ is isomorphic to the space $${\rm Lin}_2(\Bbb R^n,\Bbb R^k)$$ of $$\Bbb R^k$$-valued bilinear forms in $$\Bbb R^n$$. The isomorphism is given by $${\rm Lin}(\Bbb R^n,{\rm Lin}(\Bbb R^n,\Bbb R^k))\ni T \mapsto ((v,w)\mapsto T(v)(w))\in {\rm Lin}_2(\Bbb R^n,\Bbb R^k).$$In the one dimensional case, $$n=k=1$$, we have that $$Df(p)(h)= f'(p)h$$. So $$Df:\Bbb R \to {\rm Lin}(\Bbb R,\Bbb R)$$ is given by $$p \mapsto (h\mapsto f'(p)h)$$. Under the isomorphism $${\rm Lin}(\Bbb R^n,\Bbb R^k)\cong \Bbb R^{kn}$$ with $$k=n=1$$, $$h \mapsto f'(p)h$$ reads only $$f'(p)$$. The derivative of this is $$f''(p)$$ and, well, the matrix representing a bilinear form $$\Bbb R \times \Bbb R \to \Bbb R$$ is $$1\times 1$$, and the number inside is $$f''(p)$$ for $$D^2f(p)$$. Another way to see it is like this:$$D^2f(p)(h)(k)= \left(\frac{\rm d}{{\rm d}t}\bigg|_{t=0} Df(p+th)\right)(k) = \frac{\rm d}{{\rm d}t}\bigg|_{t=0} f'(p+th)k = f''(p)hk.$$