Search Results
| 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 real-analysis
Search options not deleted
user 2810
For questions about real analysis, such as limits, convergence of sequences, properties of the real numbers, the least upper bound property, and related analysis topics such as continuity, differentiation, and integration.
1
vote
Number of Jumps
This question is closely related to this one; in particular, if $x_k \in (a,b)$, $k=1,\ldots,n$, are such that $f(x_k+)-f(x_k-) > \varepsilon$, we have from the latter that
$$
n \varepsilon < \sum\lim …
- 24.2k
1
vote
Looking for a function $f$ that is $n$-differentiable, but $f^{(n)}$ is not continuous
Your function works for $n=1$ but not for $n=2$.
For $n=1$, the function is everywhere differentiable, and it holds
$f'_1(x) = 2x \sin(1/x) - \cos(1/x)$ for $x \neq 0$, and $f'_1 (0) = 0$; hence $f'_ …
- 24.2k
11
votes
1
answer
2k
views
How much a càdlàg (i.e., right-continuous with left limits) function can jump?
I have changed the title (replaced "well-behaved" by "càdlàg"), since it seems that "a well-behaved function" might be interpreted as "a function of bounded variation" (rather than "a càdlàg function" …
- 24.2k
8
votes
Existence of non-constant continuous functions with infinitely many zeros
It is well known that, with probability $1$, the zero set of Brownian motion is an uncountable closed set with no isolated points.
- 24.2k
0
votes
Function $\mathbb{R}\to\mathbb{R}$ that is continuous and bounded, but not uniformly continuous
As another example, consider the sequence $S_n = \sum\limits_{k = 1}^n {\frac{1}{k}} $, and a function $f$ such that $f(S_n)=1$ and $f\big(\frac{{S_n + S_{n + 1} }}{2}\big) = 0$. Further note that
…
- 24.2k
1
vote
Accepted
Prove an increasing derivative implies an increasing function
Hint: show that $h'(x)$ is positive on $(0,1)$, using the mean value theorem applied to $f(x)-f(0)$.
- 24.2k
3
votes
Accepted
a problem in real analysis
For a counterexample, just let $r(t)$ be any nonnegative continuous function on $[0,T)$ such that $\lim \inf _{t \to T} r(t) = 0$ but $\lim \sup _{t \to T} r(t) > 0$, and such that $\int_0^T {r(u)du} …
- 24.2k
3
votes
Is it possible for function $f : \mathbb{R} \to \mathbb{R}$ have a maximum at every point in...
Sample paths of Brownian motion have this property (with probability $1$), see here.
- 24.2k
2
votes
Showing range is countable
In other words, $f$ has a local minimum everywhere. More generally, the same result holds if $f$ has a local extremum everywhere; according to this (Problem 2010-4/B), the result can be found in sever …
- 24.2k
3
votes
Null Sequences and Real Analysis
Concerning the last question, it is true that, for any fixed positive integer $k$, $x_n \to 0$ if and only if $x_n^k \to 0$. Also, $x_n \to 0$ obviously implies that $x_n^n \to 0$. On the other hand, …
- 24.2k
6
votes
Accepted
Value of $\lim_{n\to\infty}{(1+\frac{2n^2+\cos{n}}{n^3+n})^n}$
Hint: Show that
$$
\lim _{n \to \infty } n\ln \bigg(1 + \frac{{2n^2 + \cos n}}{{n^3 + n}}\bigg) = 2.
$$
Elaborating:
$$
n\ln \bigg(1 + \frac{{2n^2 + \cos n}}{{n^3 + n}}\bigg) = n\ln \bigg(1 …
- 24.2k
5
votes
Accepted
Using the Mean Value Theorem to Evaluate an Integral of a Sequence of Functions?
For a concrete counterexample, consider the following.
Define $f_n$ on $[0,1/n]$ by
$$
f_n (x) = \cos ^2 \bigg(\frac{\pi }{x}\bigg), \;\; x \in [\alpha_n,1/n],
$$
and
$$
f_n (x) = 0, \;\; x \in [0 …
- 24.2k
0
votes
Convergence of Sequences (II)
Concerning the second question, as Arturo pointed out in a comment above, $\lim _{n \to \infty } a_n = \lim _{n \to \infty } b_n $. Denote the common limit by $l$. Since $a_n \leq x_n \leq b_n$ for a …
- 24.2k
3
votes
Function maximum with positive derivative
More precisely, $f$ is left differentiable at $b$, with left derivative $f'_ - (b) $ given by
$$
f'_ - (b) = \mathop {\lim }\limits_{x \to b - } \frac{{f(b) - f(x)}}{{b - x}} > 0.
$$
Put $l = f'_ …
- 24.2k
1
vote
Accepted
derivative must be bounded on interior for a differentiable function on a closed interval
Fix $x \in (a,b]$. If $f$ is differentiable at $x$, then
$$
\exists \mathop {\lim }\limits_{y \to x^ - } \frac{{f(x) - f(y)}}{{x - y}} \in \mathbb{R}.
$$
However, by the mean-value theorem
$$
\fra …
- 24.2k