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.
 A: 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 $0\leq a<b\leq 1,$ 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 $C.$
