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Let $U\subseteq \mathbb{R}^{m}$ open and simply connected, $B:U \rightarrow \mathcal{L}(\mathbb{R}^{m},\mathbb{R}^{n})$ differentiable. If $$(B^{\prime}(x). v).w= (B^{\prime}(x). w).v \ \ \ \ \ \forall \ v,w \in \mathbb{R}^{m}, x \in U$$ then there $g:U\rightarrow \mathbb{R}^{n}$ twice differentiable such that $ g ^ {\prime} (x) = B (x) $ in $ U $.

I'm thinking of doing some kind of comprehensive $ g = \int B $ but I can not clearly define this. Any suggestions are welcome.

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What is $\mathcal{L}(\mathbb{R}^{m},\mathbb{R}^{n})$? –  Keshav Srinivasan Sep 15 '13 at 3:26
    
@KeshavSrinivasan The space of linear transformations from ${\bf R}^m$ to ${\bf R}^n$. –  Pedro Tamaroff Sep 15 '13 at 3:27
    
What does $B'$ mean? The derivative of $B(x)$ w.r.t $x$? –  Shuchang Sep 15 '13 at 3:47
    
What does it mean for $B$ to be differentiable? –  Keshav Srinivasan Sep 15 '13 at 3:55
    
$B^{\prime}(x)$ means derived of $B$ in $x$ –  helmonio Sep 15 '13 at 3:58

1 Answer 1

up vote 1 down vote accepted

I'll assume that $B$ is continuously differentiable. If not, you can get the result by smoothing.

First, let's suppose that $U$ is convex, and for ease of notation, that $0 \in U$.

If there is a differentiable function $g$ on $U$ with $g' = B$, then, setting $a(t) = g(t\cdot x)$ for $t \in [0,1]$ and an arbitrary but fixed $x \in U$, we see

$$\begin{align} g(x) - g(0) &= a(1) - a(0)\\ &= \int_0^1 a'(t)\,dt\\ &= \int_0^1 g'(tx)\cdot x\,dt\\ &= \int_0^1 B(tx)\cdot x\,dt. \end{align}$$

So let's arbitrarily decide that we want $g(0)= 0$ and define

$$g(x) := \int_0^1 B(tx)\cdot x\,dt$$

for $x\in U$. We need to show that $g$ is differentiable with $g' = B$. So for $x\in U$ and $h$ small enough that also $x+h\in U$, we find

$$g(x+h) - g(x) = \int_0^1 B(t(x+h))\cdot(x+h) - B(tx)\cdot x\,dt.$$

Let $b_t(s) = B(t(x+sh))\cdot(x+sh)$, then the integrand above is $b_t(1) - b_t(0) = \int_0^1 b_t'(s)\,ds$, and

$$\begin{align} b_t'(s) &= \Bigl(B'(t(x+sh))\cdot (th)\Bigr)\cdot(x+sh) + B(t(x+sh))\cdot h\\ &= \Bigl( B'(t(x+sh))\cdot(x+sh)\Bigr)\cdot(th) + B(t(x+sh))\cdot h \end{align}$$

by the assumed symmetry of $B'$. Thus

$$\begin{align} g(x+h) - g(x) &= \int_0^1 \int_0^1 \Bigl( B'(t(x+sh))\cdot(x+sh)\Bigr)\cdot(th) + B(t(x+sh))\cdot h\,ds\,dt\\ &= \int_0^1\left(\int_0^1 tB'(t(x+sh))\cdot(x+sh) + B(t(x+sh))\,dt\right)\cdot h \,ds\\ &=\int_0^1\left(\int_0^1 \frac{d}{dt}\Bigl(t\cdot B(t(x+sh))\Bigr)\,dt\right)\cdot h \,ds\\ &= \int_0^1 B(x+sh)\cdot h\,ds\\ &= B(x)\cdot h + \underbrace{\int_0^1 B(x+sh) - B(x)\,ds}_{O(\lVert h\rVert)}\cdot h. \end{align}$$

That shows that for convex $U$ there exists a differentiable $g$ on $U$ with $g' = B$ - since $B$ itself is differentiable, $g$ is twice differentiable.

It remains to extend the result to general simply connected (open) $U$ (still assuming $0\in U$ for simplicity). Using the local (every point in $U$ has a convex neighbourhood contained in $U$) result, it follows that for any two homotopic piecewise smooth closed curves $\alpha,\beta\colon [0,1] \to U$, we have

$$\int_0^1 B(\alpha(t))\cdot \alpha'(t)\,dt = \int_0^1 B(\beta(t))\cdot \beta'(t)\,dt,$$

and since $U$ is simply connected, all closed curves in $U$ are null-homotopic, hence the integral over any (piecewise smooth) closed curve vanishes, and we can define

$$g(x) = \int_0^1 B(\gamma(t))\cdot \gamma'(t)\,dt$$

where $\gamma$ is any piecewise smooth curve in $U$ connecting $0$ and $x$. (The above says $g(x)$ does not depend on the choice of the curve, and the argument for the convex case shows that $g' = B$ on $U$.)

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thank you very much for your contribution –  helmonio Sep 17 '13 at 0:31

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