Skip to main content
Search type Search syntax
Tags [tag]
Exact "words here"
Author user:1234
user:me (yours)
Score score:3 (3+)
score:0 (none)
Answers answers:3 (3+)
answers:0 (none)
isaccepted:yes
hasaccepted:no
inquestion:1234
Views views:250
Code code:"if (foo != bar)"
Sections title:apples
body:"apples oranges"
URL url:"*.example.com"
Saves in:saves
Status closed:yes
duplicate:no
migrated:no
wiki:no
Types is:question
is:answer
Exclude -[tag]
-apples
For more details on advanced search visit our help page
Results tagged with
Search options not deleted user 26522

Real-valued functions of real variable, analytic properties of functions and sequences, limits, continuity, smoothness of these.

5 votes
0 answers
195 views

What are the possible $L^{\infty}$ closures of an integration-invariant linear subspace of $...

Let $S \subset C([0,1],\mathbb{R})$ be an $\mathbb{R}$-linear subspace that is invariant under the $T := \int_0^x$ integration operation: if $g \in S$ then the function $f = Tg$ defined pointwise by $ …
Vesselin Dimitrov's user avatar
9 votes
1 answer
493 views

Subspaces of $L^2(0,1)$ dense on every truncation $L^2(c,1)$

It may be better to move this to a separate question. Let me call a linear subspace $V \subset L^2(0,1)$ to be tame if, for every linear subspace $W \subset V$, either $W$ is dense in $L^2(0,1)$, or …
Vesselin Dimitrov's user avatar
10 votes
1 answer
581 views

Are the polynomials in $\{1/t\}$ dense in $L^2(0,1)$?

Added. My question in the title was solved (in the negative) by Nik Weaver (in the answer below) and Mateusz Kwaśnicki (in the comments). In both solutions, the reason is that the $L^2$ density fails …
Vesselin Dimitrov's user avatar
12 votes

Why could Mertens not prove the prime number theorem?

Because the scale is too small in Mertens's theorem, and the prime number theorem as well as the Riemann hypothesis are hidden by the $O(1/\log{X})$ notation. Indeed, the former amounts to strengthen …
Vesselin Dimitrov's user avatar