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Suppose $U\subseteq \mathbb{R}^{m}$ is an open and convex set, $f:U \rightarrow \mathbb{R}^{n}$ is $C^2$ such that $0 \in U$, $f(0)=0$.

Prove that $$\left \| \frac{\partial^2 f}{\partial u \,\partial v}(x) \right \| \leq \left \| u \right \|\left \| v \right \| \text{ for all }x \in U \text{ and } u,v\in \mathbb{R}^{m} $$.

I know that $\frac{\partial f}{\partial u}(x)= \nabla f(x) \cdot u$ .

Now, if I take the second directional derivative I get confuse and dont really see what the proper notation is.

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ipreferpi
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2 Answers2

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Just note that $\nabla f(x) \cdot u$ is still a function of $x$, given by $$ \nabla f(x) \cdot u = \frac{\partial f}{\partial x_1} u_1 + \frac{\partial f}{\partial x_2} u_2 $$

So now you must compute

$$ \nabla(\frac{\partial f}{\partial x_1} u_1 + \frac{\partial f}{\partial x_2} u_2)\cdot v = \frac{\partial^2 f}{\partial x_1^2}u_1 v_1 + \frac{\partial^2 f}{\partial x_1 \partial x_2} u_2 v_1 +\frac{\partial^2 f}{\partial x_2 \partial x_1}u_1 v_2 + \frac{\partial^2 f}{\partial x_2^2}u_2 v_2 $$

However, regarding the result, the data seems insufficient.

PierreCarre
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  • I forgot to add $f'(0)=0$ but since $f(0)=0$. Also i think this exercise who my prof gave isn't the correct translation (this exercise is from Lima Analisis Vol 2 Book). This notation was my main problem. Thank you – ipreferpi Apr 04 '19 at 16:56
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Note that$$\frac{\partial f}{\partial u} = \begin{pmatrix} \frac{\partial f_1}{\partial x_1} & \cdots & \frac{\partial f_1}{\partial x_m}\\ \vdots & \ddots & \vdots \\ \frac{\partial f_n}{\partial x_1}& \cdots & \frac{\partial f_n}{\partial x_m} \end{pmatrix}\begin{pmatrix} u_1\\ \vdots \\ u_m \end{pmatrix}=\begin{pmatrix} \sum_k\frac{\partial f_1}{\partial x_k}u_k\\ \vdots \\ \sum_k\frac{\partial f_n}{\partial x_k}u_k \end{pmatrix}$$so that$$\frac{\partial^2f}{\partial u\partial v}=\begin{pmatrix} \sum_k\frac{\partial^2f_1}{\partial x_k\partial x_1}u_k & \cdots & \sum_k\frac{\partial^2f_1}{\partial x_k\partial x_m}u_k\\ \vdots & \ddots & \vdots \\ \sum_k\frac{\partial^2f_n}{\partial x_k\partial x_1}u_k & \cdots & \sum_k\frac{\partial^2f_n}{\partial x_k\partial x_m}u_k \end{pmatrix}\begin{pmatrix} v_1\\ \vdots \\ v_m \end{pmatrix}=\begin{pmatrix} \sum_{k,l}\frac{\partial^2f_1}{\partial x_k\partial x_l}u_kv_l\\ \vdots \\ \sum_{k,l}\frac{\partial^2f_n}{\partial x_k\partial x_l}u_kv_l \end{pmatrix}$$and$$\left\|\frac{\partial^2f}{\partial u\partial v}(x)\right\|\leq\max_i\sum_{k,l}\left|\frac{\partial^2f_i}{\partial x_k\partial x_l}(x)\right||u_k||v_l|\leq\max_i\sum_{k,l}\left|\frac{\partial^2f_i}{\partial x_k\partial x_l}(x)\right|\|u\|\|v\|.$$

wjmolina
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