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 $x$.
-
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 $$\lim_{t\rightarrow 0}\frac{\sum\limits_{i=0}^m u_i \partial_{x_i}f(x+tu) – \sum\limits_{j=0}^m u_j \partial_{x_j}f(x)}{t}\\$$?
-
If so, how we can get to $$u_i \partial_{x_i x_j} f(x)u_j$$?
I just can't see it
Best Answer
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 $$u_i\lim_{t\rightarrow 0}\frac{\partial_{x_i}f(x+tu)-\partial_{x_i}f(x)}{t} = u_i\partial_{x_ix_j}f(x)u_j$$