# Ordinary and covariant derivative inequality: $\| u\|_1 \leq C\left( \| \frac{Du}{dt} \|_0 + \|u\|_0 \right)$

Let $$I=[0,1]$$.

Define:

$$H_0 = L^2(I,\mathbb{R}^3)$$ with inner product $$\langle u,v \rangle_0 = \int_0^1 \langle u(t), v(t) \rangle \text{ dt}$$

$$H_1 = W^{1,2}(I, \mathbb{R}^3)$$ with inner product $$\langle u,v \rangle_1 = \langle u,v \rangle_0 + \langle u',v' \rangle_0$$.

and the submanifold $$\Omega = \{\omega \in H_1\ \mid \ \|\omega(t)\| = 1 \text{ and } w(0) = w(1) \}$$ of $$H$$.

I have trouble showing the following "elementary" inequality between covariant and ordinary derivative.

If $$A$$ is an $$H_1$$ bounded subset of $$\Omega$$ then there exists a constant $$C$$ such that $$\| u\|_1 \leq C\left( \| \frac{Du}{dt} \|_0 + \|u\|_0 \right)$$ for any $$w \in A$$ and any $$u \in H_1(w^*TS^2)$$.

The covariant derivative for some $$u \in H_1(w^*TS^2)$$ along a curve $$w$$ is that of the standard metric on $$S^2$$ and can be written as $$\frac{Du}{\partial t} = u'(t) - \langle u'(t), w(t) \rangle w(t).$$

I've tried to keep the outline of the specific setting as brief and precise as possible, if there is anything unclear about it, feel free to comment.

What I've tried: Squaring both sides of the inequality and trying to get the $$\| \|_1$$ and $$\| \|_0$$ terms of $$w$$ together, since subtracting would only leave the $$L^2$$-norm of the ordinary derivative, which might help. When squaring I also used that $$\langle w(t), \frac{Du}{dt}(t) \rangle = 0$$ for any $$t$$ since $$\frac{Du}{dt}(t)$$ lies in the tangential plane $$S^2_{w(t)}$$ and is thus orthogonal to $$w(t)$$. I can't get any further though and I lack (forgot) some general things to look out for when showing such inequality.

Any help would be appreciated.

Let $$n(t)=\langle u'(t),w(t)\rangle w(t)$$ so $$u'(t)=Du/dt(t)+n(t).$$ We want to bound $$\|u\|_1^2=\|u\|_0^2+\|Du/dt\|_0^2+\|n\|_0^2$$ by $$C(\|Du\|_0+\|u\|_0)^2$$ for some $$C.$$ The only troublesome term is $$\|n\|_0^2.$$ Differentiating $$\langle u,w\rangle=0$$ gives $$n(t)=\langle u'(t),w(t)\rangle w(t)=-\langle u(t),w'(t)\rangle w(t),$$ so
$$\|n\|_0^2=\int_0^1\langle u(t),w'(t)\rangle^2dt\leq\int_0^1\| u(t)\|^2\|w'(t)\|^2dt\leq C'\|u\|_\infty^2$$ for some $$C',$$ using Cauchy-Schwarz and the assumption that $$\|w\|_1$$ is bounded. Here $$\|u\|_\infty$$ is the essential supremum of $$\|u\|.$$
Between any $$0\leq a the change in $$\|u\|^2$$ is $$\int_a^b\frac{d}{dt}\|u\|^2dt=2\int_a^b\langle u(t),u'(t)\rangle dt=2\int_a^b\langle u(t),Du/dt(t)\rangle dt.$$ To justify the first expressions use a smooth approximation in $$W^{1,2}.$$ The second equality uses the fact that $$n(t)$$ is orthogonal to $$u(t)$$ (note $$u(t)$$ lies in the tangent space to $$S^2$$ at $$w(t)$$). The right-hand-side is at most $$\|Du/dt(t)\|_0^2+\|u\|_0^2.$$ And the average value of $$\|u\|$$ is $$\int u(t)dt\leq \|u\|_0,$$ so the essential supremum of $$\|u\|$$ is at most $$C(\|u\|_0+\|Du/dt\|_0)$$ for some $$C.$$
• Thanks for the detailed input. There are some inconsistencies in your answer, however, making it harder to follow: The definition of $n(t)$ your using for your calculations seem to vary from the one at the top. Also how is $\langle u,w \rangle = 0$? It’s the covariant derivative of $u$ that is orthogonal to $w$ isn’t it? Edit: I made a mistake looking at your calculations with $n$. The second point still stands, i will check the first one now. – Nhat Jan 26 '19 at 18:32
• @Nhat: I've corrected a couple of missing $w$'s where I used $n(t),$ and added "note $u(t)$ lies in the tangent plane to $S^2$ at $w(t)$" - at least that's how I understand it – Dap Jan 26 '19 at 18:37
• I’m not fluent in differential geometry enough so you may be right. Is the tangent plane at $w(t)$ the one with $w(t)$ at the origin? Or is it the plane with origin at $0$ that’s tangent to $S^2$ after translation? I feel like i’m misunderstanding the difference between tangent space and tangent plane. – Nhat Jan 26 '19 at 18:42
• So the picture i have in mind of a curve $u \in H_1(w^*TS^2)$ being a curve that follows $w$ through its tangent planes is wrong then. What a shame - but it makes sense, I just checked the definition by the author (Curve Straightening Flow by Langer / Singer) and it says $u$ is closed at $0$ which only makes sense with your interpretation. I will check your calculations when I’m at home and then mark it as answered. Tyvm! – Nhat Jan 26 '19 at 18:47