# Equality of directional derivatives.

I am not sure how to prove: $$d_{u}d_{v}f=d_{v}d_{u}f$$ if $f$ is $C^{2}$ for any vectors $u$ and $v$. A theorem says $d_{v}f(y)=df(y)v$ but I am not sure how to apply it to solve this problem. Any help is appreciated.

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We use Einstein summation notation. Let $u = u^{i} e_{i}$ and $v = v^{i} e_{i}$. Then we have

$$d_u d_v f = u^i D_i (v^j D_j f) = u^i v^j D_{ij} f = v^j D_j (u^i D_i f) = d_v d_u f,$$

proving the claim. In a more familiar notation, let $u, v \in \mathbb{R}^{d}$ be $d$-dimensional vectors and $f : \mathbb{R}^d \to \mathbb{R}$ be of the class $C^2$. If we expand $u$ and $v$ with respect to the standard basis $\{ e_1, \cdots, e_n \}$

$$u = \sum_{i = 1}^{d} u^i e_i \quad \text{and} \quad v = \sum_{i = 1}^{d} v^i e_i,$$

then we have

$$d_u f = \sum_{i=1}^{d} u^i \frac{\partial f}{\partial x^i}.$$

Thus we have

$$d_u d_v f = \sum_{i=1}^{d} u^i \frac{\partial}{\partial x^i} \left( \sum_{j=1}^{d} v^j \frac{\partial f}{\partial x^j} \right) = \sum_{i=1}^{d} \sum_{j=1}^{d} u^i v^j \frac{\partial^2 f}{\partial x^i \partial x^j}.$$

Since the Hessian of $f$ is symmetric, or in other words

$$\frac{\partial^2 f}{\partial x^i \partial x^j} = \frac{\partial^2 f}{\partial x^j \partial x^i},$$

it follows that

$$d_u d_v f = d_v d_u f$$

as desired.

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 But what if $f\colon\mathbb{R}^{n}\to\mathbb{R}^{m}$? – MathMajor Mar 6 '12 at 0:43 If $f = (f_1, \cdots, f_m)$, then we have $d_v f = (d_v f_1, \cdots, d_v f_m)$, thus the general dimension case immediately follows. – sos440 Mar 6 '12 at 13:22