Linear Algebra - set of piecewise continuous functions is a vector space

[corey115]

Homework Statement

A function f:[a,b] $\rightarrow$ ℝ is called piecewise continuous if there exists a finite number of points a = x₀ < x₁ < x₂ < ... < x_k-1 < x_k = b such that
(a) f is continuous on (x_i-1, x_i) for i = 0, 1, 2, ..., k
(b) the one sided limits exist as finite numbers

Let V be the set of all piecewise continuous functions on [a, b]. Prove that V is a vector space over ℝ, with addition and scalar multiplication defined as usual for functions.

Homework Equations

The axioms for fields and vector spaces.

The Attempt at a Solution

If function addition and scalar multiplication work in the same way as usual, don't I just have to make an argument that the limits still exist as finite numbers (since our scalars are finite)? And a similar argument that if I add two functions that those limits will just be the sum of the two limits of each function?

 

[jbunniii]

If function addition and scalar multiplication work in the same way as usual, don't I just have to make an argument that the limits still exist as finite numbers (since our scalars are finite)? And a similar argument that if I add two functions that those limits will just be the sum of the two limits of each function?

Suppose $f, g\in V$, and let $h = f + g$. If $f$ and $g$ are both continuous at a point $x$, what can you say about $h$ at that point? Is it continuous at $x$?

 

[corey115]

Suppose $f, g\in V$, and let $h = f + g$. If $f$ and $g$ are both continuous at a point $x$, what can you say about $h$ at that point? Is it continuous at $x$?

It will also be continuous because the sum of two continuous functions will also be continuous. But do I need extra arguments besides that because V is the set of piecewise continuous functions?

 

[jbunniii]

It will also be continuous because the sum of two continuous functions will also be continuous. But do I need extra arguments besides that because V is the set of piecewise continuous functions?

Yes, but we have learned something already. $h$ will be continuous at any point where both $f$ and $g$ are continuous, so $h$ can only be discontinuous at points where $f$ or $g$ are discontinuous, and there are only finitely many such points. Thus $h$ can have only finitely many discontinuities. So $h$ satisfies condition (a).

Now consider condition (b). If $h$ is discontinuous at $x$, what are $\lim_{y \rightarrow x^+} h(x)$ and $\lim_{y \rightarrow x^-} h(x)$? Can you write these in terms of the corresponding one-sided limits of $f$ and $g$?

 

[corey115]

Yes, but we have learned something already. $h$ will be continuous at any point where both $f$ and $g$ are continuous, so $h$ can only be discontinuous at points where $f$ or $g$ are discontinuous, and there are only finitely many such points. Thus $h$ can have only finitely many discontinuities. So $h$ satisfies condition (a).

Now consider condition (b). If $h$ is discontinuous at $x$, what are $\lim_{y \rightarrow x^+} h(x)$ and $\lim_{y \rightarrow x^-} h(x)$? Can you write these in terms of the corresponding one-sided limits of $f$ and $g$?

Would that not be just the sum of the two limits of f and g at x, or if one or both of those two functions is discontinuous at x, then I would just use the whatever sided-limit that I need?

 

[jbunniii]

Would that not be just the sum of the two limits of f and g at x, or if one or both of those two functions is discontinuous at x, then I would just use the whatever sided-limit that I need?

Right, you have $\lim_{y \rightarrow x^+} h(y) = \lim_{y \rightarrow x^+} f(y) + \lim_{y \rightarrow x^+} g(y)$, and similarly with the left hand limit. Since these limits exist for $f$ and $g$, they also exist for $h$. This shows that $h$ satisfies (b).

 

[jbunniii]

So that shows that $V$ is closed under addition. What else do you need to show in order to prove that $V$ is a vector space?

 

[corey115]

So that shows that $V$ is closed under addition. What else do you need to show in order to prove that $V$ is a vector space?

Closure under scalar multiplication.
Commutative/Associative for addition/multiplication.
Additive identity/inverse.
Multiplicative identity.
Distributive property.
Associative property for scalar multiplication.

 

[jbunniii]

Commutative/Associative for addition/multiplication.
Distributive property.
Associative property for scalar multiplication.

OK, these three are true for any functions so they remain true for piecewise continuous functions.

Additive identity/inverse.
Multiplicative identity.

Which functions do these correspond to? Are they piecewise continuous?

Closure under scalar multiplication.

If $f \in V$ and $c$ is a scalar, is $cf$ piecewise continuous? Should be pretty easy to show that it is.

 

[corey115]

OK, these three are true for any functions so they remain true for piecewise continuous functions.

Which functions do these correspond to? Are they piecewise continuous?

If $f \in V$ and $c$ is a scalar, is $cf$ piecewise continuous? Should be pretty easy to show that it is.

For the 2nd item you listed, would the additive inverse of f be -f and the additive identity of f be the 0 function that is discontinuous where ever f is discontinuous?

As for the 3rd item, I'm not entirely sure how I would go about showing that. It makes sense intuitively as to why it would be true, but I'm not sure how to put that on paper.

 

[jbunniii]

For the 2nd item you listed, would the additive inverse of f be -f

Yes. Note that $-f$ is discontinuous at exactly the same points as $f$, and the left and right limits exist because they exist for $f$. (Insert easy proofs as needed.)

and the additive identity of f be the 0 function that is discontinuous where ever f is discontinuous?

The additive identity of $V$ would be just the standard 0 function, which is continuous everywhere. Note that all continuous functions are in $V$ because the number of discontinuities (zero) is certainly finite, and the left and right limits exist due to continuity.

As for the 3rd item, I'm not entirely sure how I would go about showing that. It makes sense intuitively as to why it would be true, but I'm not sure how to put that on paper.

Claim: $cf$ is continuous at any point $x$ where $f$ is continuous. (Proof?) Therefore $cf$ has at most finitely many discontinuities. What about $\lim_{y \rightarrow x^+} cf(y)$? What does this equal?

 

[corey115]

Yes. Note that $-f$ is discontinuous at exactly the same points as $f$, and the left and right limits exist because they exist for $f$. (Insert easy proofs as needed.)

The additive identity of $V$ would be just the standard 0 function, which is continuous everywhere. Note that all continuous functions are in $V$ because the number of discontinuities (zero) is certainly finite, and the left and right limits exist due to continuity.

Claim: $cf$ is continuous at any point $x$ where $f$ is continuous. (Proof?) Therefore $cf$ has at most finitely many discontinuities. What about $\lim_{y \rightarrow x^+} cf(y)$? What does this equal?

Since that is just a constant (scalar in our case), it would equal c times whatever the limit of just f(x) was.

 

[jbunniii]

Since that is just a constant (scalar in our case), it would equal c times whatever the limit of just f(x) was.

Right. Expressing it formally:
$$\lim_{y \rightarrow x^+} cf(y) = c \lim_{y \rightarrow x^+}f(y)$$
by the linearity property of limits, and the right hand side exists because $f \in V$.

 

[corey115]

Right. Expressing it formally:
$$\lim_{y \rightarrow x^+} cf(y) = c \lim_{y \rightarrow x^+}f(y)$$
by the linearity property of limits, and the right hand side exists because $f \in V$.

You have been more than helpful! I feel like I have a much greater understanding of this proof now!