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In Carmo's Differential Geometry of Curves and Surfaces, the book proves that given a differentiable function $f: \mathbb{R}^3 \to \mathbb{R}$ and $a \in \mathbb{R}$ a regular value of $f$, $f^{-1}(a)$ is an orientable surface in $\mathbb{R}^3$. The argument goes as follows: the unit gradient field $\frac{\nabla f}{|\nabla f|}$ is well defined and nonzero on $f^{-1}(a)$ since $a$ is a regular value of $f$, and is orthogonal to $f^{-1}(a)$ everywhere. It then asserts without proof that $\frac{\nabla f}{|\nabla f|}$ is differentiable on $f^{-1}(a)$, and hence satisfies the conditions for orientability. However, it's not obvious to me that this vector field is even continuous, since we only assumed that $f$ is differentiable. Of course, even if $\frac{\nabla f}{|\nabla f|}$ is not continuous over $\mathbb{R}^3$, it could be continuous over $f^{-1}(a)$ - but I don't see what forces it to be continuous or differentiable. If anyone could explain why (or give a counterexample, in the unlikely event the book is wrong), that would be great.

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    For any smooth nonvanishing vector field defined on a small ball, the normalization of it is also smooth. You can see this by what you've learnt in your multivariable calculus course. Just write down the components in local coordinates and find the formula for the normalization, and see it is indeed smooth away from where it is 0(hence normalization not defined). – 2017-02-15
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    $f$ does not need to be smooth. It is only differentiable, and differentiable does not imply continuously differentiable. – 2017-02-15
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    If you didn't even assume $C^1$ continuity, then you're right; even the gradient in and of itself is not differentiable. I think, if do carmo tries to prove in such a generality, this proof is just wrong. – 2017-02-15
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    And in that case, you'd better appeal to the connectedness argument, rather than this normal vector argument. – 2017-02-15
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    Perhaps, but two points here: 1) the normalized gradient could be continuous even if the gradient is not, and 2) continuity is required only on a very restricted subset of $\mathbb{R}^3$, namely $f^{-1}(a)$, so it seems plausible that continuity/differentiability might be provable in this context. Do you know of any counterexamples? – 2017-02-15
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    What's the connectedness argument? – 2017-02-15
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    Well, it may happen in some cases, but I don't think the normalized gradient of merely differentiable function is always continuous. (No concrete example though.) By connectedness argument I mean the surface $f^{-1}(a)$ split its complement in two pieces: the preimage of the positive ray starting at $a$ and the preimage of the negative ray starting at $a$. (It may happen that $f^{-1}(a)$ is not connected, but each connected component divides $\mathbb{R}^3$ into two pieces, so it's okay) – 2017-02-15
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    Please note that DoCarmo defines a function to be differentiable if it is $C^1$ (p. 52). So all this fuss is not needed!! – 2017-02-15
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    It actually seems that DoCarmo defines "differentiable" to be smooth according to p. 52. In any case, he definitely doesn't consider functions which are merely differentiable. – 2017-02-15
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    Ah, that would explain why he appeared to be a bit careless about issues like this. Nevertheless, I still wonder what a counterexample, if it exists, would look like; also, if orientability can be proven without assuming continuity of the gradient, a counterexample would leave us with the weird situation of having a continuous normal field which is not given by the unit gradient! – 2017-02-15
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    I'm not even sure then whether $f^{-1}(a)$ is a diffeomorphic embedding so that we can say about tangent spaces, needless to say it admits a normal vector field. – 2017-02-15
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    Well, the proof in DoCarmo that $f^{-1}(a)$ is a regular surface doesn't seem to use anything more than the differentiability of $f$ - of course, this requires that the definition of "regular surface" be weakened to require local coordinate charts to only be differentiable with an invertible derivative, but this is all you need to define the tangent plane. In this case, since the coordinate charts are no longer even $C^1$, it seems unlikely that a continuous normal vector field exists in all cases. – 2017-02-15
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    A related follow-up question: if $f^{-1}(a)$ is an orientable surface, does this imply that $\frac{\nabla f}{|\nabla f|}$ is continuous over $f^{-1}(a)$? My intuition says yes, since a counterexample would have to involve the gradient suddenly flipping directions; the fact that derivatives have the Darboux property suggests that shouldn't happen, although that fact is only for single-variable functions so it's no use if we want to prove the statement formally (if it's even true). – 2017-02-15
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    @JeffreyDawson: Note that to prove $f^{-1}(a)$ is a regular surface, one needs the Implicit Function Theorem, and for this one truly needs $C^1$ and not just differentiability. – 2017-02-15
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    @TedShifrin: I think everywhere differentiability is sufficient: see http://mathoverflow.net/questions/182030/does-the-implicit-function-theorem-hold-for-discontinuously-differentiable-funct?noredirect=1&lq=1 and http://mathoverflow.net/questions/75049/does-the-inverse-function-theorem-hold-for-everywhere-differentiable-maps – 2017-02-15

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