calculus - Is $f(x)=1/x$ continuous on $(0,\infty)$? - Mathematics ...
A function is called uniformly continuous if you can prove that given epsilon, the required value of delta depends on epsilon but NOT c. 1/x is NOT uniformly continuous on (0,1) which is why you can't get around the fact that delta depends on c. 1/x would be uniformly continuous on other intervals however (intervals where f' is bounded). Share.
What is a continuous extension? - Mathematics Stack Exchange
1. f(a) is defined. 2. lim x → af(x) exists. 3. lim x → af(x) = f(a) Suppose the #2 holds and let's consider the possibilities. If #1 also holds, then either #3 holds or #3 doesn't hold. If #3 holds, then the function is continuous at x = a. If #3 doesn't hold, then the function is not continuous at x = a, but the failure is simply due to ...
Hölder- continuous function - Mathematics Stack Exchange
8. This question already has answers here: Function on [a, b] [a, b] that satisfies a Hölder condition of order α> 1 α> 1 is constant (2 answers) Closed 7 years ago. f: I → R f: I → R is said to be Hölder continuous if ∃α> 0 ∃ α> 0 such that |f(x) − …
Difference between continuity and uniform continuity
2014年1月27日 · 84. First of all, continuity is defined at a point c, whereas uniform continuity is defined on a set A. That makes a big difference. But your interpretation is rather correct: the point c is part of the data, and is kept fixed as, for instance, f itself. Roughly speaking, uniform continuity requires the existence of a single δ> 0 that works ...
prove $x \\mapsto x^2$ is continuous - Mathematics Stack …
$\begingroup$ @VLC A little to late for you, but it may help someone else with the same problem. You want to choose your $\delta > 0$ so that it satisfies $$ \delta < \min\left\{1 , \frac{\epsilon}{1+2\lvert x_0 \rvert} \right\}.$$ Choosing $$\delta = \frac12 \min\left\{1 , \frac{\epsilon}{1+2\lvert x_0 \rvert} \right\}$$ does this, but it did not need to be with $\frac12$ …
Continuity and Joint Continuity - Mathematics Stack Exchange
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Why are norms continuous? - Mathematics Stack Exchange
11. To keep it short and straight to the point: the norm of the normed space (X, ∥ ⋅ ∥) (X, ‖ ⋅ ‖) is a continuous function because the topology you (usually) consider on X X is the smallest topology in which ∥ ⋅ ∥ ‖ ⋅ ‖ is continuous. So it is continuous because we want it to be continuous.
Show that the function $f(x, y)$ = $xy$ is continuous.
$\begingroup$ The statement "the product of two continuous function is continuous" is equivalent to "the function $(x,y)\mapsto xy$ is continuous" $\endgroup$ – Milo Brandt Commented Feb 22, 2015 at 17:17
Continuous function defined on compact set is uniformly continuous
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Continuous Poisson Distribution - Mathematics Stack Exchange
2021年8月25日 · As a consequence, the Continuous-Poisson distribution can be naturally characterized with: Continuous-Poisson(λ).lt(k) = Gamma(k).gt(λ) Continuous-Poisson (λ).lt (k) = Gamma (k).gt (λ) It's straightforward to get the CDF of our Continuous-Poisson(λ) Continuous-Poisson (λ) by plugging in the CDF of the Gamma distribution from Wikipedia as.