# Is the functional $w \mapsto \int_0^1 | \ \| w(t) \| - 1 | \ dt$ $C^1$ or even smooth?

Let $$H:= H(I;\mathbb{R}^3)$$ be the space of $$L^2$$ + absolutely continuous functions with $$L^2$$ derivative.

For $$w \in H$$ consider the functional $$\psi(w) = \int_0^1 | \ \| w(t) \|^2 - 1 | \ dt$$ where $$\| \cdot \|$$ denotes the standard norm in $$\mathbb {R^3}$$ and $$| \cdot |$$ the absolute value in $$\mathbb{R}$$.

I want to show that $$\Omega := H(I;S^2) = \psi^{-1}(0)$$ is a $$C^\infty$$-submanifold of codimension $$1$$ of $$H$$.

I think I'm at the point where I only need to make sure that the functional $$\psi$$ is smooth.

To compute the first derivative, I calculated the first variation which is $$\left.\frac{d}{d\varepsilon}\right|_{\varepsilon=0} \psi(w + \varepsilon \eta) = \int_0^1 \text{sgn}(\| w \|^2 - 1) \ 2 w\eta\ dt \ \ \ \ \ \ \text{ for all } \eta \in H$$

(which is a linear and bounded functional in $$\eta$$ and makes it the Frechét differential $$D_w \psi$$ at $$w$$ that is unequal $$0$$ making the implicit function theorem usable for the proof of the highlighted statement above.)

In turn $$\Omega$$ is at least a $$C^1$$-submanifold. Now calculating further derivatives / variations of $$\psi$$ to show smoothness, the second variation yields

$$\left.\frac{d}{d\varepsilon^2}\right|_{\varepsilon=0} \psi(w + \varepsilon \eta) = \left.\frac{d}{d\varepsilon}\right|_{\varepsilon=0} \int_0^1 \text{sgn}(\| w \|^2 + 2\varepsilon w \eta + \varepsilon^2\|\eta \|^2 - 1) \ 2(w + \varepsilon\eta)\ dt \ \ \ \ \ \$$

The derivative of the sgn-function is $$0$$ almost everywhere, continuing above terms with product rule yields (? not sure here)

$$\int_0^1 \text{sgn}(\| w \|^2 - 1) \ 2 \eta\ dt$$

Is this correct?

• $\psi(w)=0$ means that $\|w\|\equiv1$, and your differentials can't be right (because $\psi$ is not smooth due to the absolute value not being differentiable at 0.) However, you can replace the absolute value in the integral by its square, which gives you the same zero set, and then everything is nice and smooth. – Lukas Geyer Jan 27 '19 at 20:59
• @LukasGeyer Squaring is much better and makes everything smooth, thanks for the correction. I assumed that the curves being absolutely continuous would make the absolute value we have differentiale a.e. but I suppose I was wrong. One more thing: Isn’t $\| w \| = 1$ what I want though? Squaring is smoother, I see, but doesn’t it yield the same (desired) preimage for 0? – Nhat Jan 27 '19 at 21:12
• Yes, it does yield exactly the same preimage, mappings into the unit sphere. – Lukas Geyer Jan 28 '19 at 2:24
• While the preimage is indeed as desired, the first variation now contains the term $w^2 -1$ due to chain rule making the differential $0$ on $\Omega$ and the implicit function theorem unusable. I will need to figure something else out, I suppose. - @LukasGeyer Feel free to post your comment as an answer so I can close the question. – Nhat Jan 28 '19 at 11:52