Andre
Reputation
2,343
Top tag
Next privilege 2,500 Rep.
Create tag synonyms
 Nov 21 comment Maximal ideals in the algebra of continuously differentiable functions on [0,1] Okay, I see. I was confused about complex and real differentiable. Thanks. Nov 20 comment Maximal ideals in the algebra of continuously differentiable functions on [0,1] Why is $g_t \in \mathfrak a$ ? This would be true if either $\mathfrak a$ is closed under conjugation or if the conjugate of a functions is again continuously differentiable ? But conjugating does not preserve the $C^1$ property, I guess. Nov 15 answered A query in the definition of multivariable function's limit Nov 15 revised Let $\mathscr{P}_n(\Bbb{R})$ denote the vector space of polynomials with degree $\le$ n… added 65 characters in body Nov 14 answered Let $\mathscr{P}_n(\Bbb{R})$ denote the vector space of polynomials with degree $\le$ n… Nov 14 answered Two non-differentiable functions whose product is differentiable. Nov 3 awarded Popular Question Nov 2 comment Prove that operator is surjective. Sure, $X$ itself did not came into my mind, but this is certainly more precise. Nov 2 answered Prove that operator is surjective. Oct 25 comment Riemann sum of a function that is discontinuous I guess you want to compute these over a bounded interval ? Oct 8 comment Derivations having local character I now read math.toronto.edu/mgualt/MAT1300/1300%20Lecture%20notes.pdf which makes it clear. Oct 4 asked Derivations having local character Sep 30 awarded Notable Question Sep 27 comment Maximal abelian subalgebra of Banach algebra is closed and contains the unit Why should $B_c$ be a subalgebra ? A prodcut like $(\lambda c + b_1')(\mu c + b_2')$ contains the element $\lambda \mu c^2$. Sep 26 awarded Yearling Sep 8 awarded Revival Sep 2 comment rudin's definition of a compact set $K$ is assumed to be a subset of $X$, so Rudin means a cover of $K$ by opens in $X$. Aug 30 revised Show that a given set has full measure or measure 0 deleted 139 characters in body Aug 20 revised Show that a given set has full measure or measure 0 added 7 characters in body Aug 20 comment Show that a given set has full measure or measure 0 Okay Bungo is right. To get the desired result I want to have that $m((E+(x-y)) \cap (x-\epsilon,x+\epsilon)) = m(E \cap (x-\epsilon,x+\epsilon))$. I will think a moment about that. Does it follow from density of $\mathbb Q$ in $\mathbb R$ that $m(E \cap I) = m((E+z) \cap I)$ for every open set $I$ and real number $z$ ?