How does a vector-valued functions inherit properties from its components

continuityfunctionslipschitz-functionsreal-analysis

Let $f: \mathbb{R}^n \to \mathbb{R}^m$ be a vector-valued function.

My question is: how does $f$ inherit its property from its components? and if a property holds for all $f_i$, does that necessarily mean that $f$ satisfies said property as well?

For example: is it true that if all $f_i$, $i \in \{1, \ldots, n\}$, the components of $f = (f_i)$ satisfies property $P$, then $f$ satisfies property $P$ as well?

If every $f_i$ is differentiable, then $f$ is differentiable.

If every $f_i$ is continuous, then $f$ is continuous.

If every $f_i$ is Lipschitz, then $f$ is Lipschitz.

Of course, taking the intersection of the domains for which properties hold into account.

Also for properties that makes sense for vector-valued functions (this excludes things like convexity)

Does this always hold?

Best Answer

Let $\pi_i:\Bbb{R}^m\to\Bbb{R}$ be the function $(a_1,\dots, a_m)\mapsto a_i$. So, it's the $i^{th}$ coordinate mapping. Similarly, let $\theta_i:\Bbb{R}\to\Bbb{R}^m$ be defined as $\theta_i(a)=(0,\dots \underbrace{a}_{\text{$i^{th}$ spot}},\dots, 0)$. Then, we have the following obvious relations: \begin{align} f_i&= \pi_i\circ f \quad \text{and}\quad f=\sum_{i=1}^m\theta_i\circ f_i. \end{align}

So,

  • a property $P$ which is stable under composition by a (surjective) linear map means that if $f$ has that property, each $f_i$ does as well.
  • a property which is stable under composition by (injective) linear maps and also finite sums means that if each $f_i$ has that property, then so does $f$.

In your 3 examples, differentiability, continuity and Lipschitz continuity are all preserved under composition by linear maps. They are also preserved under finite sums. This is why for each of these properties, $f$ has it if and only if each $f_i$ has it.

For example, another simple property of this type is $P=\text{linearity}$. Clearly, linearity of functions is stable under compositions and finite sums, hence $f$ is linear if and only if each $f_i$ is linear.

An example of a property which is not fully preserved is injectivity, and this is obvious because $f_i=\pi_i\circ f$ is the composition of a surjective non-injective function $\pi_i$ with $f$. So, even if $f$ is bijective, there is no reason to expect $f_i$ to be injective. As a concrete counter example, take $f:\Bbb{R}^2\to\Bbb{R}^2$ to be the identity map. Then, $f_1=\pi_1\circ f=\pi_1\circ \text{id}=\pi_1$ is not injective. You can come up with many other such examples.

Related Question