0
$\begingroup$

I have been reading this answer and I couldn't get the proof.

The argument is about how we can write the second directional derivative of $f:\mathbf{R}^m\rightarrow\mathbf{R}$ with respect to direction $u$. The argument goes as follow:

First directional derivative of $f:\mathbf{R}^m\rightarrow \mathbf{R}$ in the direction of $u$ at $x$ is given by \begin{equation} \partial_u f(x):=\lim_{t\rightarrow 0}\frac{f(x+tu)-f(x)}{t}=\nabla f(x) \cdot u = \sum_{i=1}^{m} u_i\partial_{x_i}f(x). \label{} \end{equation} The second directional derivative along the direction $u$ is given in the similar fasion: \begin{align*} \partial^2_{uu}f(x)&=\partial_u(\partial_u f)\\ &=\lim_{t\rightarrow 0}\frac{\partial_u f(x+tu)-\partial_u f(x)}{t}\\ &=\lim_{t\rightarrow 0}\frac{\nabla f(x+tu)\cdot u-\nabla f(x)\cdot u}{t}\\ &=\lim_{t\rightarrow 0}\frac{u_i \partial_{x_i}f(x+tu)-u_i \partial_{x_i}f(x)}{t}\\ &=u_i \partial_{x_i x_j} f(x)u_j\\ &=u^THu \label{} \end{align*} where $H=D^2 f(x)$ is the Hessian matrix of $f$ at $x$.

  1. Is $$\lim_{t\rightarrow 0}\frac{u_i \partial_{x_i}f(x+tu)-u_i \partial_{x_i}f(x)}{t}$$ the same as $$\lim_{t\rightarrow 0}\frac{\sum\limits_{i=0}^m u_i \partial_{x_i}f(x+tu) - \sum\limits_{j=0}^m u_j \partial_{x_j}f(x)}{t}\\$$?

  2. If so, how we can get to $$u_i \partial_{x_i x_j} f(x)u_j$$?

I just can't see it

$\endgroup$
1
$\begingroup$

The answer to question one use yes $$u_i\partial_{x_i}f(x+tu) = \sum_{i=0}^m u_i\partial_{x_i}f(x+tu)$$ and it holds true even for the second factor. This is called Einstein notation, mainly: repeated indices are to be summed over.

For the second question $$\lim_{t\rightarrow 0}\frac{u_i\partial_{x_i}f(x+tu)-u_i\partial_{x_i}f(x)}{t} = u_i\underbrace{\lim_{t\rightarrow 0}\frac{\partial_{x_i}f(x+tu)-\partial_{x_i}f(x)}{t}}_{\text{dir derivative of }\partial_{x_i}f(x) \text{ along }u}$$ The second limit, being the directional derivative of the function $\partial_{x_i}f(x)$ evaluates, with some abuse of notation, to $$\nabla (\partial_{x_i}f(x))\cdot u = \partial_{x_j}(\partial_{x_i}f(x))u_j = \partial_{x_ix_j}f(x)u_j$$ where I expanded the scalar product using Einstein notation. Now plugging it back into the limit we get $$u_i\lim_{t\rightarrow 0}\frac{\partial_{x_i}f(x+tu)-\partial_{x_i}f(x)}{t} = u_i\partial_{x_ix_j}f(x)u_j$$

$\endgroup$
  • $\begingroup$ I don't think there is any matrix in my second question... $\endgroup$ – rtrtrt Aug 3 '18 at 20:18
  • $\begingroup$ Yes sorry, I'm editing! $\endgroup$ – Davide Morgante Aug 3 '18 at 20:19
  • $\begingroup$ Thanks. I have difficulties in following the Einstein notation, but by making an example with $m=2$ it is clear (in particular I didn't understand why you took $u_i$ out if the indexes of the summation are different, one is over $i$ and the other is over $j$, but now it is clear) $\endgroup$ – rtrtrt Aug 3 '18 at 20:36

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.