# Is it correct to move the $\lim$ operator inside $\| \cdot \|$ in deriving this partial derivative?

Let $$X$$ be open in $$\mathbb R^n$$, $$F$$ a Banach space, and $$m \in \mathbb N^*$$. Suppose $$f:X \to F$$ such that $$\partial_{j_1} \cdots \partial_{j_{m+1}} f$$ and $$\partial^m f$$ exist in a neighborhood of $$a$$ for all $$j_1, \ldots, j_{m+1} \in \{1,\ldots,n\}$$. Assume $$h^i = \left (h_1^i, \ldots, h_n^i\right ) \in \mathbb R^n$$ with $$1 \le i \le m$$. We define a map $$A$$ by $$\begin{array}{l|rcl} A & {(\mathbb R^n)}^m & \longrightarrow & F \\ & \left [h^1, \ldots,h^m\right ] & \longmapsto & \sum_{j_1, \ldots, j_m =1}^n \partial_j \partial_{j_1} \cdots \partial_{j_m} f (a) \left (h^1_{j_1} \cdots h^m_{j_m}\right ) \end{array}$$

I have verified that $$A$$ is a multilinear map and thus $$A \in \mathcal L^m(\mathbb R^n, F)$$. Moreover, it follows from the definition of mixed partial derivative that $$\partial^m f(a) \in L^m(\mathbb R^n, F)$$. In the following, I try to prove that $$\partial_j (\partial^m f)(a) = A$$

I'm not sure if, in (5), I correctly move the lim operator inside the norm operator. Could you please verify if this step is correct? Thank you so much!

My attempt:

First, we have \begin{aligned} &\frac{\partial^m f(a +te_j) - \partial^m f(a)}{t} \left [h^1, \ldots,h^m\right ] - A \left [h^1, \ldots,h^m\right ] \\ ={}& \frac{\partial^m f(a +te_j)[h^1, \ldots,h^m] - \partial^m f(a) [h^1, \ldots,h^m]}{t} - A [h^1, \ldots,h^m]\\ ={}& \frac{ \sum_{j_1, \ldots, j_m =1}^n \partial_{j_1} \cdots \partial_{j_m} f (a + te_j) \left (h^1_{j_1} \cdots h^m_{j_m}\right )- \sum_{j_1, \ldots, j_m =1}^n \partial_{j_1} \cdots \partial_{j_m} f (a) \left (h^1_{j_1} \cdots h^m_{j_m}\right )}{t} \\ & \quad - \sum_{j_1, \ldots, j_m =1}^n \partial_j \partial_{j_1} \cdots \partial_{j_m} f (a) \left (h^1_{j_1} \cdots h^m_{j_m}\right ) \\ ={}& \sum_{j_1, \ldots, j_m =1}^n \left (h^1_{j_1} \cdots h^m_{j_m} \right) \cdot \left ( \frac{ \partial_{j_1} \cdots \partial_{j_m} f (a + te_j) -\partial_{j_1} \cdots \partial_{j_m} f (a) }{t} - \partial_j \partial_{j_1} \cdots \partial_{j_m} f (a) \right ) \end{aligned}

It follows that

\begin{aligned} & \lim_{t \to 0} \left \| \frac{\partial^m f(a +te_j) - \partial^m f(a)}{t} - A\right \| \\ \overset{(1)}{=}{}& \lim_{t \to 0} \sup_{\|h^1\|_1\le1,\ldots,\|h^m\|_1 \le 1} \left \| \sum_{j_1, \ldots, j_m =1}^n \frac{\partial^m f(a +te_j) - \partial^m f(a)}{t} \left [h^1, \ldots,h^m\right ] - A \left [h^1, \ldots,h^m\right ] \right \|\\ \overset{(2)}{\le}{}& \lim_{t \to 0} \sup_{\|h^1\|_1\le 1,\ldots,\|h^m\|_1 \le 1} \sum_{j_1, \ldots, j_m =1}^n \left |h^1_{j_1} \cdots h^m_{j_m} \right | \cdot\bigg \| \frac{ \partial_{j_1} \cdots \partial_{j_m} f (a + te_j) - \partial_{j_1} \cdots \partial_{j_m} f (a) }{t} \\ \overset{(3)}{\le}{}& \lim_{t \to 0} \sup_{\|h^1\|_1\le 1,\ldots,\|h^m\|_1 \le 1} \sum_{j_1, \ldots, j_m =1}^n \|h^1\|_1 \cdots \|h^m\|_1 \cdot\bigg \| \frac{ \partial_{j_1} \cdots \partial_{j_m} f (a + te_j) - \partial_{j_1} \cdots \partial_{j_m} f (a) }{t} \\ & \quad - \partial_j \partial_{j_1} \cdots \partial_{j_m} f (a)\bigg \|\\ \overset{(4)}{\le}{}& \lim_{t \to 0} \sum_{j_1, \ldots, j_m =1}^n \left \| \frac{ \partial_{j_1} \cdots \partial_{j_m} f (a + te_j) - \partial_{j_1} \cdots \partial_{j_m} f (a) }{t} - \partial_j \partial_{j_1} \cdots \partial_{j_m} f (a)\right \|\\ ={}& \color{blue}{\sum_{j_1, \ldots, j_m =1}^n \lim_{t \to 0} \left \| \frac{ \partial_{j_1} \cdots \partial_{j_m} f (a + te_j) - \partial_{j_1} \cdots \partial_{j_m} f (a) }{t} - \partial_j \partial_{j_1} \cdots \partial_{j_m} f (a)\right \|}\\ \overset{(5)}{=}{}& \color{blue}{\sum_{j_1, \ldots, j_m =1}^n \left \| \lim_{t \to 0} \frac{ \partial_{j_1} \cdots \partial_{j_m} f (a + te_j) - \partial_{j_1} \cdots \partial_{j_m} f (a) }{t} - \partial_j \partial_{j_1} \cdots \partial_{j_m} f (a)\right \|}\\ ={}& \sum_{j_1, \ldots, j_m =1}^n \left \| \partial_j \left ( \partial_{j_1} \cdots \partial_{j_m} f \right ) (a) - \partial_j \partial_{j_1} \cdots \partial_{j_m} f (a) \right \|\\ ={}& \sum_{j_1, \ldots, j_m =1}^n \left \| \partial_j \partial_{j_1} \cdots \partial_{j_m} f (a) - \partial_j \partial_{j_1} \cdots \partial_{j_m} f (a) \right \|\\ ={}& 0 \end{aligned}

$$(1)$$: This follows from the definition of the operator norm of a multilinear map.

$$(2)$$: This follows from triangle inequality.

$$(3)$$: It follows from the definition of $$\|\cdot\|_1$$ that $$|h^1_j| \le \|h^1\|_1,\ldots, |h^m_j| \le \|h^m\|_1$$ for all $$j \in \{1,\ldots,n\}$$. As such, $$\left |h^1_{j_1} \cdots h^m_{j_m} \right | =\left |h^1_{j_1}\right | \cdots \left | h^m_{j_m} \right | \le \|h^1\|_1 \cdots \|h^m\|_1$$.

$$(4)$$: It follows from $$\|h^1\|_1\le 1,\ldots,\|h^m\|_1 \le 1$$ that $$\|h^1\|_1 \cdots \|h^m\|_1 \le 1$$.

Hence $$\partial_j (\partial^m f)(a) = A$$

• This is correct because the norm of any normed vector spaced is continuous. Commented Nov 23, 2019 at 9:08
• Hi @MaximilianJanisch , I would like to confirm that even the map $$\frac{ \partial_{j_1} \cdots \partial_{j_m} f (a + te_j) - \partial_{j_1} \cdots \partial_{j_m} f (a) }{t}$$ is not defined at $t=0$, we still have the equality [...] Commented Nov 23, 2019 at 9:23
• [...] $$\lim_{t \to 0} \left \| \frac{ \partial_{j_1} \cdots \partial_{j_m} f (a + te_j) - \partial_{j_1} \cdots \partial_{j_m} f (a) }{t} - \partial_j \partial_{j_1} \cdots \partial_{j_m} f (a)\right \| = \left \| \lim_{t \to 0} \frac{ \partial_{j_1} \cdots \partial_{j_m} f (a + te_j) - \partial_{j_1} \cdots \partial_{j_m} f (a) }{t} - \partial_j \partial_{j_1} \cdots \partial_{j_m} f (a)\right \|$$, right? Commented Nov 23, 2019 at 9:23
• As long as the limit is well-defined there should be no problem Commented Nov 23, 2019 at 9:24
• Thank you so much @MaximilianJanisch :) Commented Nov 23, 2019 at 9:25