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 Apr17 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 $[0,1]$. Apr17 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. Apr17 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. Apr17 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 $||D(f)-D(g)||_\infty$. Mar15 comment Floor Function Homomorphism and Isomorphism The group operation here is addition, not multiplication. Mar12 awarded Popular Question Mar12 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 $5v_1 - 2v_2 + 3v_3 = 0$ Feb14 comment “Length” of an element in a free group You are looking for the word metric on a group, I believe. Feb12 comment Apparent counterexample to the Picard-Lindelöf theorem What are the assumptions of the Picard-Lindelof theorem? Does $xy^{1/2}$ satisfy those assumptions? Feb3 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 $y \cdot z = x$?) Feb3 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. Feb3 answered Prove Inverses not need be Unique if Associativity fails Jan31 awarded Informed Jan20 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 $[a,b]$. Jan20 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. Jan17 answered Proving that Doob's martingale is a martingale Jan16 comment Proving that Doob's martingale is a martingale It might also help to specify what exactly is $I_n$ in your question. Jan16 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. Jan3 awarded Enthusiast Dec25 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...