# How do you efficiently show that for the vector space V where f(x) is continuous on [0,15] and W where f(x) is differentiable on [0,15], W is a subspace in V, i.e. W sube V?

## I already made an attempt at this, and I know from calculus that if $f \left(x\right)$ is differentiable, then $f \left(x\right)$ is continuous. But maybe I'm overthinking it, because I'm not sure if my proof is sufficient. I would like to be able to do this with less effort, because I think perhaps I'm being redundant in my proof. Here was my attempt at this: $W = \left\{f \left(x\right) | \exists f ' \left(x\right) \forall x \in \left[0 , 15\right]\right\}$ $V = \left\{f \left(x\right) | \exists f \left(x\right) \forall x \in \left[0 , 15\right]\right\}$ 1) $\exists f \left(x\right) = \sin x$ such that $f ' \left(x\right)$ exists since $f ' \left(x\right) = \cos x$; $\cos x \in \mathbb{R} , \therefore \cos x \in \left[0 , 15\right]$. $\therefore W \notin \emptyset$ 2) Let $c$ be a scalar $\in \mathbb{R}$. Then, $c f \left(x\right) = f \left(c x\right)$. $\left(c f \left(x\right)\right) ' = c \left(f \left(x\right)\right) ' = c f ' \left(x\right)$ $f ' \left(x\right)$ exists $\forall x \in \left[0 , 15\right]$, so $c f ' \left(x\right)$ exists for the same conditions and thus $\left(c f \left(x\right)\right) '$ exists $\forall x \in \left[0 , 15\right]$ $\therefore$ since $f \left(x\right)$ must not have any discontinuities, cusps, corners, etc. to be differentiable, there exist no conditions for which $f \left(x\right)$ is not continuous. Thus $f ' \left(x\right)$ is continuous. 3) Let there be an $f \left(x\right)$ and $g \left(x\right)$ such that $\left[f \left(x\right)\right] ' = f ' \left(x\right)$ and $\left[g \left(x\right)\right] ' = g ' \left(x\right)$. $\left[\left(f + g\right) \left(x\right)\right] ' = f ' \left(x\right) + g ' \left(x\right) = \left(f ' + g '\right) \left(x\right)$ $f \left(x\right)$ and $g \left(x\right)$ are both differentiable, and $\left[\left(f + g\right) \left(x\right)\right] ' = \left(f ' + g '\right) \left(x\right)$. Therefore, $W \subseteq V$.

Mar 18, 2016

While most of the ideas are correct, there are a few small problems with that proof. First, $V$ is not correctly defined. $\left\{f \left(x\right) | \exists f \left(x\right) \forall x \in \left[0 , 15\right]\right\}$ is simply the set of all functions defined on $\left[0 , 15\right]$. The should look something more like $\left\{f : \mathbb{R} \to \mathbb{R} | \forall x \in \left[0 , 15\right] , f \text{ is continuous at } x\right\}$

The notation $W \subseteq V$ simply implies that $W$ is a subset of $V$, and not necessarily a subspace. This fact is simply shown as differentiability implies continuity.

The second and third parts don't seem to be stating the desired result, that is, that $W$ is closed under vector addition and scalar multiplication. What follows is a slightly modified version of the given proof:

As a function is differentiable at a point only if it is continuous at that point, we have that $f \in W$ only if $f \in V$, meaning $W \subseteq V$.

1) As $f \left(x\right) = \sin \left(x\right)$ is differentiable on $\left[0 , 15\right]$, we have $\sin \left(x\right) \in W$, and therefore $W \ne \emptyset$.

2) Take any scalar $c$. Then for any differentiable function $f \in W$, we have $\left(c f\right) ' \left(x\right) = c f ' \left(x\right)$ at each point in $\left[0 , 15\right]$. Thus $c f$ is also differentiable on $\left[0 , 15\right]$ and therefore $c f \in W$, meaning $W$ is closed under scalar multiplication.

3) Take any $f , g \in W$. Then $\left(f + g\right) ' \left(x\right) = f ' \left(x\right) + g ' \left(x\right)$ at each $x$ in $\left[0 , 15\right]$. Thus $\left(f + g\right)$ is differentiable on $\left[0 , 15\right]$ and therefore $\left(f + g\right) \in W$, meaning $W$ is closed under vector addition.

As $W$ is a nonempty subset of $V$ which is closed under the inherited operations of vector addition and scalar multiplication, $W$ is a subspace of $V$.