why arent non continuous functions in an interval a 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?
why does f+g have to be continuous? I can see that even for discontinuous functions they are closed under addition and multiplication
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.
If it were linear, it would mean that f + g would be discontinuous if f and g are. i dont understand this part
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.
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