Linear space

  • Thread starter batballbat
  • Start date
  • #1
127
0
why arent non continuous functions in an interval a linear space?
 

Answers and Replies

  • #2
CompuChip
Science Advisor
Homework Helper
4,302
47
Why don't you try to find a counterexample?

Hint: try looking at closure under addition: can you find two discontinuous functions f and g such that f + g is continuous?
 
  • #3
127
0
why does f+g have to be continuous? I can see that even for discontinuous functions they are closed under addition and multiplication
 
  • #4
Deveno
Science Advisor
906
6
suppose our interval is [0,1]. define f:[0,1]→R by:

f(x) = -1, for 0 ≤ x < 1/2
f(x) = 1, for 1/2 ≤ x ≤ 1.

clearly, f is discontinuous (at 1/2).

now define g:[0,1]→R by:

g(x) = 1 for 0 ≤ x < 1/2
gx) = -1, for 1/2 ≤ x ≤ 1.

again, g(x) is discontinuous (at 1/2).

but (f+g)(x) = 0, for all x in [0,1], and constant functions are continuous.
 
  • #5
HallsofIvy
Science Advisor
Homework Helper
41,833
962
In other words, "discontinuous functions" are NOT closed under addition. For a counter example to closure under multiplication, let f(x)= 1 if x is rational, -1 if x is irrational and let g(x)= -1 if x is rational, 1 if x is irrational.
 
  • #6
CompuChip
Science Advisor
Homework Helper
4,302
47
why does f+g have to be continuous?
Because you wanted to show that the space of discontinuous is not linear. If it were linear, it would mean that f + g would be discontinuous if f and g are.
 
  • #7
127
0
If it were linear, it would mean that f + g would be discontinuous if f and g are. i dont understand this part
 
  • #8
22,089
3,296
By definition, a linear space V must satisfy: if x and y are in V, so is x+y.

If V would be the set of discontinuous functions, then the above becomes: if x and y are discontinuous functions, so is x+y.

The above example shows that this is false, hence the discontinuous functions do not form a linear space.
 
  • #10
HallsofIvy
Science Advisor
Homework Helper
41,833
962
Or, perhaps much more simply, every linear space must contain an additive identity. Since the addition here is ordinary addition of functions, the "additive identity" is the 0 function (f(x)= 0 for all x). That is a continuous function and so is NOT in the set of discontinuous functions.
 

Related Threads on Linear space

  • Last Post
Replies
7
Views
1K
  • Last Post
Replies
19
Views
3K
  • Last Post
Replies
14
Views
2K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
2
Views
2K
Replies
3
Views
2K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
8
Views
1K
Replies
4
Views
1K
Top