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Linear Algebra  set of piecewise continuous functions is a vector space 
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#1
Jan2513, 11:19 AM

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1. The problem statement, all variables and given/known data
A function f:[a,b] [itex]\rightarrow[/itex] ℝ is called piecewise continuous if there exists a finite number of points a = x_{0} < x_{1} < x_{2} < ... < x_{k1} < x_{k} = b such that (a) f is continuous on (x_{i1}, 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. 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? 


#2
Jan2513, 11:30 AM

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#3
Jan2513, 11:41 AM

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#4
Jan2513, 11:45 AM

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Linear Algebra  set of piecewise continuous functions is a vector space
Now consider condition (b). If [itex]h[/itex] is discontinuous at [itex]x[/itex], what are [itex]\lim_{y \rightarrow x^+} h(x)[/itex] and [itex]\lim_{y \rightarrow x^} h(x)[/itex]? Can you write these in terms of the corresponding onesided limits of [itex]f[/itex] and [itex]g[/itex]? 


#5
Jan2513, 11:59 AM

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#6
Jan2513, 12:04 PM

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#7
Jan2513, 12:06 PM

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



#8
Jan2513, 12:11 PM

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


#9
Jan2513, 12:29 PM

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#10
Jan2513, 01:08 PM

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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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Jan2513, 01:30 PM

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#12
Jan2513, 04:38 PM

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#13
Jan2513, 04:46 PM

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$$\lim_{y \rightarrow x^+} cf(y) = c \lim_{y \rightarrow x^+}f(y)$$ by the linearity property of limits, and the right hand side exists because [itex]f \in V[/itex]. 


#14
Jan2513, 05:21 PM

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