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

by corey115
Tags: algebra, continuous, functions, linear, piecewise, space, vector
 P: 8 1. The problem statement, all variables and given/known data 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. 2. Relevant equations The axioms for fields and vector spaces. 3. 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?
HW Helper
PF Gold
P: 2,897
 Quote by corey115 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 $x$?
P: 8
 Quote by jbunniii 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?

HW Helper
PF Gold
P: 2,897

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

 Quote by corey115 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 $g$?
P: 8
 Quote by jbunniii 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?
HW Helper
PF Gold
P: 2,897
 Quote by corey115 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).
 Sci Advisor HW Helper PF Gold P: 2,897 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?
P: 8
 Quote by 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?
Closure under scalar multiplication.
Multiplicative identity.
Distributive property.
Associative property for scalar multiplication.
HW Helper
PF Gold
P: 2,897
 Quote by corey115 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.
P: 8
 Quote by jbunniii 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.
HW Helper
PF Gold
P: 2,897
 Quote by corey115 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?
P: 8
 Quote by jbunniii 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.
HW Helper
PF Gold
P: 2,897
 Quote by corey115 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 $f \in V$.
P: 8
 Quote by jbunniii 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!

 Related Discussions Linear & Abstract Algebra 3 Calculus & Beyond Homework 14 Calculus 0 Calculus & Beyond Homework 5 Calculus 4