# 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#?

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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# .

I already made an attempt at this, and I know from calculus that if

Here was my attempt at this:

1)

2) Let

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

3) Let there be an

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

Therefore,

##### 1 Answer

While most of the ideas are correct, there are a few small problems with that proof. First,

The notation

The second and third parts don't seem to be stating the desired result, that is, that

As a function is differentiable at a point only if it is continuous at that point, we have that

1) As

2) Take any scalar

3) Take any

As