Question

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(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`.

Answer 1

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:

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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`.