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If $(V, \langle \cdot, \cdot \rangle)$ is a finite-dimensional inner product space and $f,g : \mathbb{R} \longrightarrow V$ are differentiable functions, a straightforward calculation with components shows that

$$ \frac{d}{dt} \langle f, g \rangle = \langle f(t), g^{\prime}(t) \rangle + \langle f^{\prime}(t), g(t) \rangle $$

This approach is not very satisfying. However, attempting to apply the definition of the derivative directly doesn't seem to work for me. Is there a slick, perhaps intrinsic way, to prove this that doesn't involve working in coordinates?

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2 Answers 2

up vote 14 down vote accepted

Observe that $$ \begin{align*} \frac{1}{h} & \left[ \langle f(t+h),\, g(t+h)\rangle - \langle f(t),\, g(t) \rangle \right] \\ & = \frac{1}{h} \left[ \langle f(t+h),\, g(t+h)\rangle - \langle f(t),\, g(t+h)\rangle \right] + \frac{1}{h} \left[ \langle f(t),\, g(t+h)\rangle - \langle f(t),\, g(t)\rangle \right] \\ &= \left\langle \frac{1}{h} \left[ f(t+h) - f(t) \right],\, g(t+h) \right\rangle + \left\langle f(t),\, \frac{1}{h} \left[ g(t+h) - g(t) \right] \right\rangle. \end{align*} $$ As $h\to 0$ the first expression converges to $$ \frac{d}{dt} \langle f(t), g(t) \rangle $$ and the last expression converges to $$ \langle f^{\prime}(t), g(t) \rangle + \langle f(t), g^{\prime}(t) \rangle $$ by definition of the derivative, by continuity of $g$ and by continuity of the scalar product. Hence the desired equality follows.

Note that this doesn't use finite-dimensionality and that the argument is the exact same as the one for the ordinary product rule from calculus.

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Thanks Theo, very nice answer –  ItsNotObvious Jan 4 '12 at 16:30

This answer may be needlessly complicated if you don't want such generality, taking the approach of first finding the Fréchet derivative of a bilinear operator.

If $V$, $W$, and $Z$ are normed spaces, and if $T:V\times W\to Z$ is a continuous (real) bilinear operator, meaning that there exists $C\geq 0$ such that $\|T(v,w)\|\leq C\|v\|\|w\|$ for all $v\in V$ and $w\in W$, then the derivative of $T$ at $(v_0,w_0)$ is $DT|_{(v_0,w_0)}(v,w)=T(v,w_0)+T(v_0,w)$. (I am assuming that $V\times W$ is given a norm equivalent with $\|(v,w)\|=\sqrt{\|v\|^2+\|w\|^2}$.) This follows from the straightforward computation

$$\frac{\|T(v_0+v,w_0+w)-T(v_0,w_0)-(T(v,w_0)+T(v_0,w))\|}{\|(v,w)\|}=\frac{\|T(v,w)\|}{\|(v,w)\|}\leq C\frac{\|v\|\|w\|}{\|(v,w)\|}\to 0$$

as $(v,w)\to 0$.

With $V=W$, $Z=\mathbb R$ or $Z=\mathbb C$, and $T:V\times V\to Z$ the inner product, this gives $DT_{(v_0,w_0)}(v,w)=\langle v,w_0\rangle+\langle v_0,w\rangle$. Now if $f,g:\mathbb R\to V$ are differentiable, then $F:\mathbb R\to V\times V$ defined by $F(t)=(f(t),g(t))$ is differentiable with $DF|_t(h)=h(f'(t),g'(t))$. By the chain rule,

$$D(T\circ F)|_{t}(h) =DT|_{F(t)}\circ DF|_t(h)=h(\langle f'(t),g(t)\rangle+\langle f(t),g'(t)\rangle),$$

which means $\frac{d}{dt} \langle f, g \rangle = \langle f'(t),g(t)\rangle+\langle f(t),g'(t)\rangle$.

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This is also a great answer. –  ItsNotObvious Jan 4 '12 at 16:31

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