$\langle f(t),g(t)\rangle' = \langle f'(t),g(t)\rangle + \langle f(t),g'(t)\rangle$ for differentiable $f,g : \mathcal{R} \to \mathcal{R}^n$?

Question

Suppose that $f,g$ are differentiable functions from $\mathcal{R}$ to $\mathcal{R}^n$.

Show that $\langle f(t),g(t)\rangle' = \langle f'(t),g(t)\rangle + \langle f(t),g'(t)\rangle$

I've been banging my head against a brick wall on this for a while, I can see that the RHS is something like the product rule for differentiation. We're not given an inner product but even just trying to by parts integrate the standard inner wasn't working.

Thanks in advance for any help.

Answer 1

Write $f(t)=(f_1(t),f_2(t),\dots,f_n(t))$ and $g(t)=(g_1(t),g_2(t),\dots,g_n(t))$. The inner product is given by $$\langle f(t),g(t)\rangle=\sum_{i=1}^nf_i(t)g_i(t).$$ You apply sum rule and product rule of differentiation to see that $$\frac{d}{dt}\langle f(t),g(t)\rangle=\sum_{i=1}^nf_i'(t)g_i(t)+f_i(t)g_i'(t).$$ The last expression is equal to $$\langle f'(t),g(t)\rangle + \langle f(t),g'(t)\rangle.$$

There is also a general formula telling us how to calculate derivatives of bilinear functions, but you may not yet be ready to understand. Given a bilinear function $B$ that takes a pair of vectors $(x,y)$ as input, the derivative at $(a,b)$ is the linear map $B'(a,b)$ that acts on $(x,y)$ by the formula $$B'(a,b)(x,y)=B(a,y)+B(x,b).$$

Your map $t\mapsto\langle f(t),g(t)\rangle$ is the composition $t\mapsto(f(t),g(t))$, $(x,y)\mapsto\langle x,y\rangle$. So alternatively, you can apply chain rule to obtain the formula for its derivative.

Answer 2

The proof is very similar to how you prove that $(fg)' = f'g + fg'$ for any functions $f,g\colon \Bbb R\to\Bbb R$. To wit, write \begin{align*} \langle f(t),g(t)\rangle' &= \lim_{h\to0}\frac{\langle f(t+h),g(t+h)\rangle-\langle f(t),g(t)\rangle}{h}. \end{align*} Let me write $fg$, i.e. just the concatenation, to mean $\langle f,g\rangle$. There shouldn't be any confusion by doing this. With this notation, we have \begin{align*} \langle f(t),g(t)\rangle' &= \lim_{h\to0}\frac{f(t+h)g(t+h)-f(t)g(t)}{h} \\ &= \lim_{h\to0}\frac{f(t+h)g(t+h)-f(t)g(t+h)+f(t)g(t+h)-f(t)g(t)}{h} \\ &= \lim_{h\to0}\frac{f(t+h)-f(t)}{h}g(t+h) + f(t)\lim_{h\to0}\frac{g(t+h)-g(t)}{h}\\ &= f'(t)g(t) + f(t)g'(t) \\ &= \langle f'(t),g(t)\rangle + \langle f(t),g'(t)\rangle, \end{align*} where we switched back to the usual notation in the last line. In the second-to-last equality, we used continuity of the inner product $\langle\cdot,\cdot\rangle\colon \Bbb R^n\times\Bbb R^n\to\Bbb R$.

Answer 3

A slightly different approach.

Let $f\in C^1(\Bbb R^n\times \Bbb R^m,\Bbb R^\ell)$, then it is easy to check that

$$\partial f(x,y)(a,b)=D_x f(x,y)a+D_y f(x,y)b\tag1$$

where $\partial f$ is the Fréchet derivative of $f$ and $D_x f$ is the Fréchet derivative of $f(\cdot,y)$. Similarly $D_y f$ is the Fréchet derivative of $f(x,\cdot)$.

Let $d:\Bbb R^n\times \Bbb R^n\to\Bbb R$ a dot product, hence

$$\partial\, d(x,y)(a,b)=D_x d(x,y)a+D_y d(x,y)b\tag2$$

Now note that the functions $d(\cdot ,y)$ and $d(x,\cdot)$ are linear, so

$$D_x d(x,y)=d(\cdot, y)\implies D_x d(x,y)a=d(a,y)\tag3$$

And similarly $D_y d(x,y)b=d(x,b)$. Putting all together we find that

$$\partial\, d(x,y)(a,b)=d(a,y)+d(x,b)\tag4$$

By last, using the chain rule, we have that

$$\partial\, [d(f,g)]=\partial d(f,g)\partial(f,g)=\partial d(f,g)(f',g')\\=d(f',g)+d(f,g')\tag5$$

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The key points here are $(1)$ and the fact that if $A$ is a linear function then $\partial Ax=A$, what can be checked easily using the definition of Fréchet derivative.