Vector-Valued Derivatives

Complex Functions

In general, the definition of a Derivative holds for complex functions defined on $[a,b]$.

• Theorem (Differentiable is Continuous)
• Theorem (Derivative Rules) also hold (with the same proofs). For complex function $f$, the real $f_1$ and imaginary $f_2$ parts can be described as:

$$f(t)=f_1(t)+i{f}_2(t)$$

for $t\in [a,b]$ and as $f_1,f_2$ are real,

$$f^{\prime }(x)=f_1^{\prime }(x)+i{f}_2^{\prime }(x)$$

$f$ is differentiable at $x$ iff $f_1,f_2$ are both differentiable at $x$.

(Weak) Mean Value Theorem Fails

The Theorem (Lagrange Mean Value Theorem) does not hold. For some real $x$,

$$f(x)=e^{ix}=\cos x+i\sin x$$

Then $f(2\pi )=f(0)=1-1=0$, but $f^{\prime }(x)=i{e}^{ix}$ and $|f^{\prime }(x)|=1$ for all real $x$. So,

$$1=f^{\prime }(x)=\frac{f(2\pi )-f(0)}{2\pi -0}=0$$

which is a contradiction.

L’ Hopital’s Rule Fails

On the segment $(0,1)$, let $f(x)=x$ and

$$g(x)=x+x^2{e}^{i/x^2}$$

Since $|e^{it}|=1$ for all real $t$, we see that

$$\underset{x\to 0}{\lim }\frac{f(x)}{g(x)}=1$$

But then

$$g^{\prime }(x)=1+\left(2x-\frac{2i}{x}\right)e^{i/x^2}$$

for $x\in (0,1)$ and that

$$|g^{\prime }(x)|\ge \left|2x-\frac{2i}{x}\right|-1\ge \frac{2}{x}-1=\frac{2-x}{x}$$

Hence

$$\left|\frac{f^{\prime }(x)}{g^{\prime }(x)}\right|=\frac{1}{|g^{\prime }(x)|}\le \frac{x}{2-x}$$

and so

$$\underset{x\to 0}{\lim }\frac{f^{\prime }(x)}{g^{\prime }(x)}=0$$

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

From Theorem (Lagrange Mean Value Theorem) we have

$$|f(b)-f(a)|\le (b-a)\underset{a<x<b}{\sup }|f^{\prime }(x)|$$

Theorem (Vector MVT)

This is a stronger form since it shows the existence of a point.

Suppose $\text{f}$ is a continuous mapping of $[a,b]$ into ${\mathbb{R}}^k$ and $\text{f}$ is differentiable in $(a,b)$. Then there exists $x\in (a,b)$ such that

$$|\text{vvf}(b)-\text{vvf}(a)|\le (b-a)|{\text{vvf}}^{\prime }(x)|$$