Linear space

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?

 

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.

 

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.

 

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.

 

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.

 

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.