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

 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?
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P: 3,288
 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?

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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?
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 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: 3,288 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.
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 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.
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.
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 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.
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 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!

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