Is the set of even functions in C([-1,1],R) closed and dense in C([-1,1],R)?

[benjamin111]

Homework Statement

Let Ce([0,1], R) be the set of even functions in C([0,1], R), show that Ce is closed and not dense in C.

Homework Equations

The Attempt at a Solution

I think I can solve this if I can show that even functions converge to even functions, but I can't quite figure out how to go about doing this...

 

[quasar987]

This question does not make sense to me.

A function is "even" is f(t)=f(-t). Here, for any t in (0,1], -t is out of [0,1] and thus f(-t) is not even defined.

What do you mean by even?

 

[Dick]

What does C([0,1],R) mean?

 

[quasar987]

That would be the space of continuous functions on [0,1] no doubt.

But do you see what it means for a fct to be even in this setting?

 

[Dick]

I was wondering if it could mean continuous functions from [0,1]xR->R with 'even' meaning f(x,y)=f(x,-y). The failure of the question to make any obvious sense otherwise was giving me doubts.

 

[benjamin111]

I'm sorry all. I meant that the space is [-1,1] rather than [0,1]. Sorry again and do appreciate any help.

 

[Dick]

I'm sorry all. I meant that the space is [-1,1] rather than [0,1]. Sorry again and do appreciate any help.

Then it's super easy. Take a sequence of even functions f_i converging to a function f. Take any point x, then f_i(x)->f(x) and f_i(-x)->f(-x). Need I say more?