Vector-Valued Derivatives

Definition (Vector-Valued Functions)

The vector-valued $\text{f}$ maps $[a,b]\to {\mathbb{R}}^k$. The same definition holds for ${\text{f}}^{\prime }(x)$ which is now

$$\underset{t\to x}{\lim }\left|\frac{\text{f}(t)-\text{f}(x)}{t-x}-{\text{f}}^{\prime }(x)\right|=0$$

and ${\text{f}}^{\prime }$ is another function with values in ${\mathbb{R}}^k$. We have

$$\text{f}=(f_1,f_2,\cdots ,f_k)$$

is differentiable at $x$ iff its components $f_i$ are differentiable.

Modified Vector MVT

Assume $\text{f}:[a,b]\to {\mathbb{R}}^k$ is continuous and ${\text{f}}^{\prime }$ exists on $(a,b)$, then similar to Lagrange MVT, we have

$$\begin{array}{rl}|\text{f}(b)-\text{f}(a)|& \le (b-a)\cdot |{\text{f}}^{\prime }(c)|\\ & \le (b-a)\underset{a<x<b}{\sup }|{\text{f}}^{\prime }(c)|\end{array}$$

Proof: Construct a function

$$g:[a,b]\to {\mathbb{R}}^{k}g(x)=(\text{f}(b)-\text{f}(a))\cdot \text{f}(x)$$

Then by taking the difference:

$$\begin{array}{rl}g(b)-g(a)& =(\text{f}(b)-\text{f}(a))\cdot (\text{f}(b)-\text{a})\\ & =|\text{f}(b)-\text{f}(a){|}^2\end{array}$$

And $g$ is differentiable with $g^{\prime }=(\text{f}(b)-\text{f}(a))\cdot {\text{f}}^{\prime }(x)$. We apply MVT, such that

$$\begin{array}{r}\frac{g(b)-g(a)}{b-a}=g^{\prime }(c)c\in (a,b)\end{array}$$

So,

$$|\text{f}(b)-\text{f}(a){|}^2=\left[(\text{f}(b)-\text{f}(a))\cdot {\text{f}}^{\prime }(c)\right]\cdot (b-a)$$

Then by cancelling, by Cauchy-Schwarz , we get

$$|\text{f}(b)-\text{f}(a)|\le |{\text{f}}^{\prime }(c)|\cdot (b-a)$$

Counterexample:

Consider

$$\text{f}(t)=(-\sin t,\cos t)$$

where we cannot expect

$$\frac{\text{f}(b)-\text{f}(a)}{b-a}={\text{f}}^{\prime }(c)$$

to be true all the time, since $|\text{f}(t)|=1$. The proof is similar to Lagrange Mean Value Theorem Fails.