1
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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
2
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.