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I want to solve the following exercise.

Let A be an $\mathbb{R}$-algebra. A $derivation$ on A is a $\mathbb{R}$-linear map $D: A \to A$ obeying the Leibniz rule $$ D(ab) = D(a)b + aD(b) $$ for all $a,b \in A$.

Now let M be a manifold. Show that the algebra $$ C(M) = \{ f: M \to \mathbb{R} : f \textrm{ continuous } \} $$ has no non-trivial derivations, i.e. for every linear map $D : C(M) \to C(M)$ obeying the Leibniz rule, it follows that $D = 0$.

Hint: Use that every $f \ge 0$ can be written as a square.

I have some difficulties in understanding. I am currently reading about manifolds and tangent spaces, and I read that the tangent space could be defined as the space of all point derivations $D : C^{\infty}(M) \to \mathbb{R}$, but this space is more than just the trivial derivation $D = 0$, does it make a huge difference if I am considering smooth vs. just continuous functions? And in the exercise, how does the hint help me? I can write every $f \ge 0$ as $(\sqrt{f})^2$, and then $$ D(g^2) = 2gD(g) = gD(g+g) $$ with $g = \sqrt{f}$, does it follow that $D = 0$?.

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    Hint: For fixed $x \in M$, you can assume that $f(x) = 0$ because if $f_1 = f - f(x)$, then $df_1 = df$. Now use that $df(x) = 2 g(x) dg(x)$.2012-08-17
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    @MichaelJoyce That trick seems to fail, because if $f\geq 0$, it is not true that $f_1\geq 0$. I think it can be fixed, though.2012-08-17
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    @ThomasAndrews: The fix should be to write $f_1 = h_1 - h_2$ with $h_i$ both continuous, both $\geq 0$, and such that $h_i(x) = 0$ for both. Then argue as above that $dh_1(x) = dh_2(x) = 0$, so $df(x)=0$. This is essentially the same argument that allowed us to assume $f \geq 0$ in the first place.2012-08-17

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