# Some calculation details in elliptic PDE operator.

Define function $f$: $\mathbb R^N\to \mathbb R$ by $$f(\xi):=A\xi\cdot\xi$$ where $A$ is $N \times N$ uniformly elliptic matrix, i.e., $A\xi\cdot\xi\geq \theta\lvert\xi\lvert^2$ for some $\theta>0$.

My question is as following:

$(1)$: is function $f$ convex, or even strict convex? I think yes, and I show it by proving function $$a(t):=f(\xi+t\xi_2)$$ is convex at $0$ by computing second derivative. But I was wondering is there a better explanation by computing from vector respect?

$(2)$: What is $$\nabla f(\xi)?$$ I think it is $2A\xi$ but I can only show it by using brute force calculation, i.e., write done every element of $A\xi\cdot\xi$ and do $1-D$ calculation and guess what is should be in the vector form... Again...Is there a quicker way to do the derivative?

• What does the notation $\partial/\partial \xi$ mean? Is it the product of the 1-D operators $\partial/\partial \xi_i$? – guest Jan 2 '15 at 2:13
• Sorry, it is an typo. Corrected. – spatially Jan 2 '15 at 2:13

1) You can use the definition of convexity and proceed the same way as you would for proving the convexity of the function $x^2$: (I will use $x$ instead of $\xi$ because it saves typing). Pick $s,t\in (0,1)$ with $s+t=1$.
$$f(sx_1+tx_2) = s^2Ax_1\cdot x_1 +stAx_1\cdot x_2 + st Ax_2\cdot x_1 +t^2Ax_2\cdot x_2$$ $$= sAx_1\cdot x_1+tAx_2\cdot x_2 +(s^2-s)Ax_1\cdot x_1 +stAx_1\cdot x_2 + st Ax_2\cdot x_1 +(t^2-t)Ax_2\cdot x_2$$ $$= sAx_1\cdot x_1+tAx_2\cdot x_2-(stAx_1\cdot x_1+stAx_1\cdot x_2+stAx_2\cdot x_1 + st Ax_2\cdot x_2)$$ (because $s^2-s=t^2-t=st$) $$= sAx_1\cdot x_1+tAx_2\cdot x_2-st(Ax_1\cdot x_1+Ax_1\cdot x_2 + Ax_2\cdot x_1+Ax_2\cdot x_2)$$ $$= sAx_1\cdot x_1+tAx_2\cdot x_2 -stA(x_1+x_2)\cdot(x_1+x_2).$$ Now use the uniform ellipticity to bound the above as $$f(sx_1+tx_2)< sAx_1\cdot x_1+tAx_2\cdot x_2 -st\frac{\theta}{2} \|x_1+x_2\|^2$$ giving $$f(sx_1+tx_2)< sAx_1\cdot x_1+tAx_2\cdot x_2 =sf(x_1)+tf(x_2).$$
Fix $x$ and pick any direction vector $y$. Then in the direction of $y$ we have the derivative
$$Df_x(y) = \lim_{t\to 0}\frac{f(x+ty)-f(x)}{t}$$ $$= \lim_{t\to 0}\frac{Ax\cdot x + tAx\cdot y +tAy\cdot x + t^2Ay\cdot y-Ax\cdot x}{t}$$ $$= Ax\cdot y+Ay\cdot x.$$ Specializing in the direction of $x$ gives you what you want.