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Apr
17
comment Contractions and finding Fixed Points
Feel free to read more about fixed points here. I am sensing some confusion here about the fixed point; note that $D$ is an operator defined on the function space $C[0,1]$, not on the interval $[0,1]$. Therefore it acts on functions $f$, and so its fixed point is a function, not a point in the interval $[0,1]$.
Apr
17
comment Contractions and finding Fixed Points
By definition, a fixed point $f$ of the contraction satisfies $D(f) = f$. Apply the definition of $D$ to figure out what $f$ must be. For example, since $Df(x) = x - 2/3$ on $[2/3,1]$, we must have that $f(x) = x - 2/3$ on the same interval also.
Apr
17
comment Contractions and finding Fixed Points
Well, it looks like my first comment has the gist of it. Just note that on $[2/3,1]$, $Df$ and $Dg$ are equal, hence $|Df(x)-Dg(x)| = 0$, which finishes the proof that $D$ is a contraction. You have a sign error in your last equation, hence the confusion.
Apr
17
comment Contractions and finding Fixed Points
Please clean up the fractions in your post; it's very difficult to understand right now. Assuming that your post accurately defines $D(f)$ on the last third of $[0,1]$, then $D(f)$ and $D(g)$ are in fact equal on that interval, and so their difference on $[2/3,1]$ will not contribute anything towards the sup norm $||D(f)-D(g)||_\infty$.
Mar
15
comment Floor Function Homomorphism and Isomorphism
The group operation here is addition, not multiplication.
Mar
12
awarded  Popular Question
Mar
12
comment Show that $\lambda$=1 as eigenvalue, find one corresponding eigenvector
Your matrix multiplication is incorrect; for example, the first row of your last set of equations should be $5v_1 - 2v_2 + 3v_3 = 0$
Feb
14
comment “Length” of an element in a free group
You are looking for the word metric on a group, I believe.
Feb
12
comment Apparent counterexample to the Picard-Lindelöf theorem
What are the assumptions of the Picard-Lindelof theorem? Does $xy^{1/2}$ satisfy those assumptions?
Feb
3
comment Prove Inverses not need be Unique if Associativity fails
You should at least leave the multiplication table in, so that the multiplication results I use in my answer don't look so magical :) (how did I know that $y \cdot z = x$?)
Feb
3
comment Prove Inverses not need be Unique if Associativity fails
It looks like the point of this exercise was precisely to provide you with an example where "inverses need not be unique if associativity fails". But understandably, it's hard to state it in those terms if you don't know what exactly the exercise was meant to teach you - and of course, you can't be expected to figure out the point of the exercise without having solved it! Feel free to update your question title, but don't worry about it too much.
Feb
3
answered Prove Inverses not need be Unique if Associativity fails
Jan
31
awarded  Informed
Jan
20
comment Evaluate $ \int_{a}^{b}(A - f(x))dx$ where $A = [1/(b-a)] \cdot \int_a^b f(x)\,dx$
Pretty much. I'd add the remark that the quantity $A$ as defined in the problem is the "average" value of $f$ on the interval $[a,b]$, roughly speaking. That's why the integral of $A$ over $[a,b]$ is precisely the integral of $f$ over $[a,b]$.
Jan
20
comment If G is finite group with even number of elements and has identity e, there is a in G such that a*a=e
You might look at Cauchy's theorem for a generalization of the result you're trying to prove.
Jan
17
answered Proving that Doob's martingale is a martingale
Jan
16
comment Proving that Doob's martingale is a martingale
It might also help to specify what exactly is $I_n$ in your question.
Jan
16
comment Proving that Doob's martingale is a martingale
$E(E(Y|I_n) | I_m) = E(Y|I_m)$ holds because of the law of iterated expectations, for which $ E(E(Y|I_n)) = E(Y)$ is just a special case.
Jan
3
awarded  Enthusiast
Dec
25
comment How can I prove that every term in this sequence is postive?
Suppose for contradiction that there existed a term which was negative. Then since the sequence is decreasing...