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

• corey115
In summary: If f \in V and c is a scalar, is cf piecewise continuous? Should be pretty easy to show that it is, since c is just a scalar multiplication of f by itself.
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 < ... < xk-1 < xk = b such that
(a) f is continuous on (xi-1, xi) 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?

corey115 said:
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 $x$?

jbunniii said:
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?

corey115 said:
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 $g$?

jbunniii said:
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?

corey115 said:
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).

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?

jbunniii said:
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.
Multiplicative identity.
Distributive property.
Associative property for scalar multiplication.

corey115 said:
Distributive property.
Associative property for scalar multiplication.

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

jbunniii said:
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.

corey115 said:
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?

jbunniii said:
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.

corey115 said:
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 $f \in V$.

jbunniii said:
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!

## 1. What is a vector space?

A vector space is a mathematical structure that consists of a set of objects (called vectors) that can be added together and multiplied by scalars. It follows a set of axioms, such as closure under addition and scalar multiplication, associativity, commutativity, and distributivity, among others.

## 2. How does a set of piecewise continuous functions qualify as a vector space?

In order for a set of functions to qualify as a vector space, it must fulfill the same axioms as any other vector space. This means that the set of piecewise continuous functions must be closed under addition and scalar multiplication, and must also satisfy the other axioms such as associativity, commutativity, and distributivity.

## 3. What is meant by "piecewise continuous" in this context?

Piecewise continuous refers to a function that is continuous in different intervals, but may have discontinuities at certain points. For example, a piecewise continuous function may be continuous on the interval [0, 1], but have a discontinuity at x = 1.

## 4. Can you give an example of a piecewise continuous function that is not a vector space?

Yes, a function that is defined on a finite interval and has a jump discontinuity at one of the endpoints would not qualify as a vector space. For example, the function f(x) = 1, defined on the interval [0, 1], would not be a vector space because it does not satisfy the closure under addition axiom (since adding two different values of x would result in a constant value of 2 rather than a continuous function).

## 5. How is linear algebra used in the study of piecewise continuous functions?

Linear algebra provides a framework for analyzing and manipulating vector spaces, including sets of piecewise continuous functions. It allows us to apply techniques such as matrix operations and vector transformations to study these functions and their properties. Additionally, linear algebra is also used to solve systems of equations that arise in the study of piecewise continuous functions.

Replies
13
Views
457
Replies
2
Views
756
Replies
1
Views
872
Replies
4
Views
3K
Replies
3
Views
2K
Replies
8
Views
1K
Replies
17
Views
3K
Replies
10
Views
1K
Replies
2
Views
3K
Replies
18
Views
2K