The Vector Subspace of Real-Valued Continuous Functions

Recall on the Further Examples of Vector Spaces page that the set of real-valued function $F(-\infty, \infty)$ forms a vector space.

Now consider the subset of $F(-\infty, \infty)$ that contains the set of real-valued continuous functions. We denote this set by $C(-\infty, \infty)$. We will now prove that $C(-\infty, \infty) \subset F(-\infty, \infty)$ is a vector subspace.

Let $f, g \in C(-\infty, \infty)$ be two continuous real-valued functions, and let $a \in \mathbb{F}$.

• 9. The function $f + g$ produces a continuous function since $f$ and $g$ both have no points of discontinuities, and so $(f + g) \in C(-\infty, \infty)$, or in other words, $C(-\infty, \infty)$ is closed under addition.

• 10. The function $af$ also produces a continuous function since $f$ has no points of discontinuity and multiplication by the scalar $a$ results in either a vertical stretch or compression and does not create any discontinuities, and so $(af) \in C(-\infty, \infty)$ and so $C(-\infty, \infty)$ is closed under scalar multiplication.

Therefore the set of real-valued continuous functions is a vector subspace of the set of real-valued functions.

The Vector Subspace of Real-Valued Continuous Differentiable Functions

Denote the set $C^{(n)} (-\infty, \infty)$ to be the set of real-valued continuous and differentiable functions. Recall that if a function is differentiable then it is continuous, but not all continuous functions are differentiable, for example, the function $f(x) = \mid x \mid$ is continuous for all $x$, but is not differentiable at $x = 0$, and so $C^{(1)}(-\infty, \infty) \subset C(-\infty, \infty)$. We will now verify that $C^{(1)}(-\infty, \infty)$ is a vector subspace.

Let $f, g \in C^{(1)}(-\infty, \infty)$ be two continuous real-valued differentiable functions and let $a \in \mathbb{F}$.

• 9. $f + g$ produces a continuous differentiable function since the sum of two differentiable functions is differentiable by the sum rule for derivatives, that is $\frac{d}{dx} f(x) + \frac{d}{dx} g(x) = \frac{d}{dx} (f(x) + g(x))$, and so $(f + g) \in C^{(1)}(-\infty, \infty)$ and so $C^{(1)}(-\infty, \infty)$ is closed under addition.

• 10. $af$ also produces a continuous differentiable function implied by the scalar multiple rule for derivatives, that is $\frac{d}{dx} af(x) = a \frac{d}{dx} f(x)$ and so $(af) \in C^{(1)}(-\infty, \infty)$ and so $C^{(1)}(-\infty, \infty)$ is closed under scalar multiplication.

Therefore the set of real-valued continuous differentiable functions is a subspace of the set of real-valued continuous functions.

Note: The notation $C^{(n)}(-\infty, \infty)$ represents the set of real-valued continuous functions that are $n$-times differentiable. It is possible that show that $C^{(n)}(-\infty, \infty)$ is also a vector subspace, namely that $C^{(n)}(-\infty, \infty) \subset C^{(n-1)}(-\infty, \infty) \subset C^{(n-2)}(-\infty, \infty) \subset ... \subset C^{(2)}(-\infty, \infty) \subset C^{(1)}(-\infty, \infty)$.