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#?

Question

I already made an attempt at this, and I know from calculus that if #f(x)# is differentiable, then #f(x)# 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 = {f(x) | EE f'(x) AA x in [0,15]}#

#V = {f(x) | EE f(x) AA x in [0,15]}#

1) #EE f(x) = sinx# such that #f'(x)# exists since #f'(x) = cosx#; #cosx in RR, :. cosx in [0,15]#.

#:. W !in O/#

2) Let #c# be a scalar #in RR#. Then, #cf(x) = f(cx)#.

#(cf(x))' = c(f(x))' = cf'(x)#

#f'(x)# exists #AA x in [0,15]#, so #cf'(x)# exists for the same conditions and thus #(cf(x))'# exists #AA x in [0,15]#

#:.# since #f(x)# must not have any discontinuities, cusps, corners, etc. to be differentiable, there exist no conditions for which #f(x)# is not continuous. Thus #f'(x)# is continuous.

3) Let there be an #f(x)# and #g(x)# such that #[f(x)]' = f'(x)# and #[g(x)]' = g'(x)#.

#[(f+g)(x)]' = f'(x) + g'(x) = (f'+g')(x)#

#f(x)# and #g(x)# are both differentiable, and #[(f+g)(x)]' = (f'+g')(x)#.

Therefore, #W sube V#.

Abigail Allen · Accepted Answer

While most of the ideas are correct, there are a few small problems with that proof. First, #V# is not correctly defined. #{f(x)|EEf(x)AAx in[0,15]}# is simply the set of all functions defined on #[0,15]#. The should look something more like #{f:RR->RR|AAx in[0,15], f" is continuous at "x}# The notation #W sube 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 sube V#. 1) As #f(x) = sin(x)# is differentiable on #[0,15]#, we have #sin(x) in W#, and therefore #W!=O/#. 2) Take any scalar #c#. Then for any differentiable function #f in W#, we have #(cf)'(x) = cf'(x)# at each point in #[0,15]#. Thus #cf# is also differentiable on #[0,15]# and therefore #cf in W#, meaning #W# is closed under scalar multiplication. 3) Take any #f, g in W#. Then #(f+g)'(x) = f'(x) + g'(x)# at each #x# in #[0,15]#. Thus #(f+g)# is differentiable on #[0,15]# and therefore #(f+g) 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#.

Eli Assouline · Answer

undefined To show that $ W $ is a subspace of $ V $, where $ V $ is the vector space of functions continuous on $[0,15]$ and $ W $ is the vector space of functions differentiable on $[0,15]$, we need to prove three properties:

1. Closure under vector addition: For any $ f(x), g(x) \in W $, $ f(x) + g(x) $ is differentiable on $[0,15]$.

2. Closure under scalar multiplication: For any $ c \in \mathbb{R} $ and $ f(x) \in W $, $ cf(x) $ is differentiable on $[0,15]$.

3. Contains the zero vector: The zero function, $ f(x) = 0 $, is differentiable on $[0,15]$ and thus belongs to $ W $.

These three properties confirm that $ W $ is a subspace of $ V $.