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.

  • $\begingroup$ 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. $\endgroup$
    – Mogu
    Dec 26 '20 at 10:42

Let's focus on getting domains and codomains right, and assume that all derivatives exist, for whatever we need to.

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:$$D^2f(p)(h)(k)= \left(\frac{\rm d}{{\rm d}t}\bigg|_{t=0} Df(p+th)\right)(k) = \frac{\rm d}{{\rm d}t}\bigg|_{t=0} f'(p+th)k = f''(p)hk.$$


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