Vector space

- gorica

Vector space

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- chestercat

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- amistre64

the anticipation is killing me!!

- gorica

Let \[V=\{f:(0,2) \rightarrow R| f\ is\ continuous\ and\ \lim_{x \rightarrow o^+}f(x)=\lim_{x \rightarrow 2^-}f(x)=0 \}\]Show that \[(V,+, \cdot, R)\]is vector space.
I know I have to show thatfor all x,y in V, x+y is in V and that for all x in V and all α in R, αx is in V.

- amistre64

hmm, so something of a visual like this maybe?
|dw:1377792310168:dw|

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## More answers

- gorica

|dw:1377792442827:dw|or like this? I don't think it's much about how does them look like.

- amistre64

ive got abstract algebra group stuff cluttering my head at the moment ...

- amistre64

right, f is some general function with the stated properties
what do you recall about the requirments to define a vector space? and subspace?

- amistre64

identity, inverse, and zero is what im thinking of, but those relate to groups ...

- gorica

What I wrote above is actually for subspace. I think I could solve this if somebody could tell me definition for \[\lim_{x \rightarrow a^+} f(x)=0\] and \[\lim_{x \rightarrow a^-}f(x)=0\]

- amistre64

limit from the left and right

- amistre64

|dw:1377792859241:dw|

- amistre64

in other words:
as x approaches 2 from the left, the value approaches 0
as x approaches 0 from the right, the value approaches 0

- amistre64

it could just as well be the function: x(x-2)

- gorica

That is clear to me, but I need some definition as \[(\forall \epsilon > 0)(\exists \delta>0(\forall x_1,x_2 \in [0,1])(|x_1-x_2|<\delta \rightarrow |f(x_1)-f(x_2)|<\epsilon)\] is for continuity of f

- gorica

I mean, I need it in this form so I can use it

- anonymous

This link has some info on one-sided limit definitions near the bottom of the page: http://tutorial.math.lamar.edu/Classes/CalcI/DefnOfLimit.aspx

- amistre64

Pauls site? yeah, thats always a good bet

- anonymous

By the way @gorica, do you have to show it's a vector space or a subspace? Because what you mentioned has to do with showing a space is a subspace.

- gorica

I have to show that it is vector space.

- terenzreignz

Do we actually have to go through the nitty-gritty definition of a limit?
Why not just use
\[\Large \lim_{x\rightarrow a^+}f(x) = L \quad \implies \quad \lim_{x\rightarrow a^+}cf(x) = cL\]

- terenzreignz

And of course, the other bit,
\[\Large \lim_{x\rightarrow a^+}f(x) = L \quad \& \quad \lim_{x\rightarrow a^+}g(x) = M\]\[\Large \implies \lim_{x\rightarrow a^+}[f(x) +g(x)]= L+M\]

- gorica

That's enough? If V is a subset of some vector space I can show only those two. But I don't know if V is a subset of some vector space. Do you maybe know? Or I have to show all properties of vector space?

- terenzreignz

Let's start with basics (unless you've already done these)
Suppose f and g are two elements of V.
Then show that f+g is also in V... is this already done?

- gorica

If it is enough to write only what you wrote, it is done. I thought to try to show it by definition.

- terenzreignz

Overkill, don't you think?
I mean... does your linear algebra instructor actually want to use the Real-Analysis limit?
With the epsilon and delta? D:
...sadistic

- gorica

Not linear algebra... Metric and normed spaces :D

- terenzreignz

Oh... crud.
Then... enjoy the \(\varepsilon\) and \(\delta\) then XD
So... have you proved this?
\[\Large \lim_{x\rightarrow a^+}f(x) = L \quad \implies \quad \lim_{x\rightarrow a^+}cf(x) = cL\]

- gorica

Thanks for encouraging me :D
I haven't.

- terenzreignz

Well... okay, let's do it...

- terenzreignz

Assume that \[\Large \lim_{x\rightarrow a^+}f(x) = L\]

- terenzreignz

Then, for all \(\large \varepsilon>0\) there exists a \(\large \delta>0\) such that whenever\[\Large x -a < \delta\] then \[\Large \left|f(x) - L\right|<\Large \frac{\varepsilon}{c}\]
Catch me so far?

- gorica

I think I do :D You skipped few steps but I got it :D

- terenzreignz

Actually it's wrong, c could be negative... this is more accurate:
\[\Large \left|f(x) - L\right| < \frac{\varepsilon}{|c|}\]

- terenzreignz

So that we could multiply both sides by |c| giving
\[\Large |c|\left|f(x) - L\right| < \varepsilon\]\[\Large \left|cf(x) - cL\right| < \varepsilon\]

- terenzreignz

Thus proving that \[\Large \lim_{x\rightarrow a^+}cf(x) = cL\]

- terenzreignz

Got it? ^_^

- gorica

got it :)

- terenzreignz

Now, let's do the other one, the sum...
Suppose we have \[\Large \lim_{x\rightarrow a^+}f(x) = L\]and\[\Large \lim_{x\rightarrow a^+}g(x) = M\]

- terenzreignz

Then we begin the proof...

- terenzreignz

Let \(\large \varepsilon >0\)
Then there exist \(\large \delta_1,\delta_2>0\) such that \[\Large x-a < \delta_1 \implies \left|f(x)-L\right|<\varepsilon\]\[\Large x-a < \delta_2 \implies \left|g(x) -M\right|<\varepsilon\]
Catch me so far?

- gorica

yes, go on

- terenzreignz

Wait, neater this way...\[\Large x-a < \delta_1 \implies \left|f(x)-L\right|<\frac{\varepsilon}2\]\[\Large x-a < \delta_2 \implies \left|g(x) -M\right|<\frac{\varepsilon}2\]

- terenzreignz

After all, \(\Large \frac{\varepsilon}2\) is just as arbitrary.

- terenzreignz

Anyway, choose \(\large \delta = \min\left\{\delta_1,\delta_2\right\} \)

- terenzreignz

So that whenever \(\Large x-a < \delta\) then both \[\Large \left|f(x) -L\right|<\frac \varepsilon 2\] and \[\Large \left|g(x)-M\right|<\frac \varepsilon 2\]hold

- terenzreignz

Now, we can simply add these two inequalities...
\[\Large \left|f(x)-L\right|+\left|g(x)-M\right| < \varepsilon\]

- terenzreignz

Do you see where this is going?

- terenzreignz

TRIANGLE INEQUALITY.
|a + b| \(\le\) |a| + |b|

- gorica

Yes, it's ok, I've got it when I saw it. \[|f(x)+g(x)-(L+M)|<\epsilon\]

- terenzreignz

That's good ^_^
And THAT... proves it :D

- terenzreignz

So you now have pretty much proved the only two limit-properties that you need to prove that V is vector space... unless you also want to prove that the sum of two continuous functions is continuous? :3

- gorica

No, I think it would be enough to write it by words :D But tell me, is it enough to show only what you showed now or I have to show all properties of a vector space? Or you know that V is a subset of some vector space?

- terenzreignz

That really depends on your instructor... -.-

- amistre64

nice job .. i personally never know how far to take a proof. When do we stop trying to reinvent the wheel after all?

- terenzreignz

^_^

- gorica

ok, I hope she will be happy when she sees a lot of epsilons and deltas so she won't ask for more :D Thank you very much :)

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