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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?

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

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 $$ f(sx_1+tx_2)< sAx_1\cdot x_1+tAx_2\cdot x_2 =sf(x_1)+tf(x_2).$$

2) For the second part, in order to avoid coordinates simply use the definition and bilinearity:

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.

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