# Total derivative of inner product.

Let $$F:\Bbb R^n\times\Bbb R^n\to\Bbb R$$ be the function $F(x,y)=\langle Ax,y\rangle$ where $\langle , \rangle$ denotes the standard inner product on $\Bbb R^n$ and $A$ be an $n\times n$ real matrix. If $D$ denotes the total derivative. which of the fallowing is correct?

1). $(DF(x,y))(u,v)=\langle Au,y\rangle+\langle Ax,v\rangle$

2). $(DF(x,y))(0,0)=(0,0)$

3). $DF(x,y)$ may not exist for some $(x,y)\in\Bbb R^n\times\Bbb R^n$

4). $DF(x,y)$ doesnot exist at $(x,y)=(0,0)$

I don't know, how the total derivative is defined for Inner product. So, tell me the definition for total derivative on inner product space, also tell, How to proceed further to solve this problem? Thank you

The inner product is bilinear, so you would expect that the derivative exists. The derivative exists if we can find a linear map $B:\mathbb{R}^{n} \times \mathbb{R}^n \to \mathbb{R}$ such that $$\lvert \langle A(x+h),(y+k) \rangle - \langle Ax,y \rangle - B(h,k) \rvert = o(\lvert h\rvert+\lvert k\rvert)$$ as $h,k \to 0$. Expanding out the inner product, we find $$\langle A(x+h),(y+k) \rangle - \langle Ax,y \rangle = \langle Ah,y \rangle + \langle Ax,k \rangle + \langle Ah,k \rangle$$ Then the absolute value in the definition can only be $o(\lvert h\rvert+\lvert k\rvert)$ when the first two terms are cancelled out (notice that $h=|h| (h/|h|)$ and so on). Therefore take $$B(h,k) = \langle Ah,y \rangle + \langle Ax,k \rangle.$$ Now all you have to do is show whether the remainder of the left-hand side satisfies $$\lvert \langle Ah,k \rangle \rvert = o(\lvert h\rvert+\lvert k\rvert),$$ which you can do in the same manner as showing that $\langle Ah,y \rangle=O(h)$.
• @@ Chappers) Sir I could not understand that how $\lvert \langle Ah,k \rangle \rvert = o(\lvert h\rvert+\lvert k\rvert)$ ? Please explain.... – Empty Aug 25 '15 at 3:48
• @S.Panja-1729 By the Cauchy-Schwarz inequality, $\lvert \langle Ah,k \rangle \rvert \leqslant |Ah| \, |k|$. Then $|Ah|<K|h|$ for some $K>0$ since $A$ is a finite-dimensional matrix, so it has a maximum entry, and then you can bound explicitly the sums in the product $(Ah)_i = \sum_j A_{ij} h_j$. So you have $\lvert \langle Ah,k \rangle \rvert \leqslant K|h| \, |k|$. This last term is $o(|h|+|k|)$ by the definition: $|h| \, |k|/(|h|+|k|) \to 0$ as $(h,k) \to (0,0)$, which you can check along any path tending to $(0,0)$. – Chappers Aug 25 '15 at 13:08
• @Chappers's the symbol $O$ used here for what? Is it is for order of function i.e growth rate of function? – Akash Patalwanshi Oct 13 '17 at 2:44
• @AkashPatalwanshi It's en.wikipedia.org/wiki/Big_O_notation , applied as $h,k \to 0$, as noted after the first displayed equation. – Chappers Oct 13 '17 at 11:48