What is Uniform convergence: Definition and 162 Discussions
In the mathematical field of analysis, uniform convergence is a mode of convergence of functions stronger than pointwise convergence. A sequence of functions
(
f
n
)
{\displaystyle (f_{n})}
converges uniformly to a limiting function
f
{\displaystyle f}
on a set
E
{\displaystyle E}
if, given any arbitrarily small positive number
ϵ
{\displaystyle \epsilon }
, a number
N
{\displaystyle N}
can be found such that each of the functions
f
N
,
f
N
+
1
,
f
N
+
2
,
…
{\displaystyle f_{N},f_{N+1},f_{N+2},\ldots }
differ from
f
{\displaystyle f}
by no more than
ϵ
{\displaystyle \epsilon }
at every point
x
{\displaystyle x}
in
E
{\displaystyle E}
. Described in an informal way, if
f
n
{\displaystyle f_{n}}
converges to
f
{\displaystyle f}
uniformly, then the rate at which
f
n
(
x
)
{\displaystyle f_{n}(x)}
approaches
f
(
x
)
{\displaystyle f(x)}
is "uniform" throughout its domain in the following sense: in order to guarantee that
f
n
(
x
)
{\displaystyle f_{n}(x)}
falls within a certain distance
ϵ
{\displaystyle \epsilon }
of
f
(
x
)
{\displaystyle f(x)}
, we do not need to know the value of
x
∈
E
{\displaystyle x\in E}
in question — there can be found a single value of
N
=
N
(
ϵ
)
{\displaystyle N=N(\epsilon )}
independent of
x
{\displaystyle x}
, such that choosing
n
≥
N
{\displaystyle n\geq N}
will ensure that
f
n
(
x
)
{\displaystyle f_{n}(x)}
is within
ϵ
{\displaystyle \epsilon }
of
f
(
x
)
{\displaystyle f(x)}
for all
x
∈
E
{\displaystyle x\in E}
. In contrast, pointwise convergence of
f
n
{\displaystyle f_{n}}
to
f
{\displaystyle f}
merely guarantees that for any
x
∈
E
{\displaystyle x\in E}
given in advance, we can find
N
=
N
(
ϵ
,
x
)
{\displaystyle N=N(\epsilon ,x)}
(
N
{\displaystyle N}
can depend on the value of
x
{\displaystyle x}
) so that, for that particular
x
{\displaystyle x}
,
f
n
(
x
)
{\displaystyle f_{n}(x)}
falls within
ϵ
{\displaystyle \epsilon }
of
f
(
x
)
{\displaystyle f(x)}
whenever
n
≥
N
{\displaystyle n\geq N}
.
The difference between uniform convergence and pointwise convergence was not fully appreciated early in the history of calculus, leading to instances of faulty reasoning. The concept, which was first formalized by Karl Weierstrass, is important because several properties of the functions
f
n
{\displaystyle f_{n}}
, such as continuity, Riemann integrability, and, with additional hypotheses, differentiability, are transferred to the limit
f
{\displaystyle f}
if the convergence is uniform, but not necessarily if the convergence is not uniform.
Hi,
I'm studying for finals and I just need some feedback.
One of questions MIGHT be:
If I know a sequence of functions is pointwise convergent, how do I show that it's not uniformly convergent?
I think that a pointwise convergent sequence of functions might not converge to a continuous...
This question arised in my last math class:
If a sequence of functions f_n uniformly converges to some f on (a, b) (bounded) and all f_n are integrable on (a, b), does this imply that f is also integrable on (a, b) ??
(f_n do not necessarily have to be continous, if they were, the answer...
"Suppose f_n are defined and continuous on an interval I. Assume that f_n converges uniformly to f on I. If a_n in I is a sequence and a_n -> a, prove that f_n(a_n) converges to f(a)."
I don't understand the question. Doesn't uniform convergence imply that for all x in I and e>0, | f_n(x) -...
Hello,
I have two questions to ask regarding uniform convergence for sequences of functions.
So I know that if a sequence of continuous functions f_n : [a,b] -> R converge uniformly to function f, then f is continuous.
Is this true if "continous" is replaced with "piecewise continuous"...
does Fn(x)= nx(1 - x^2)^n converge uniformly on [0,1]?
my first instinct was yes it converges uniformly to 0
but I can't seem to show that using the definition.
i get |nx(1 - x^2)^n|<=|nx|<=n for x in [0,1]
any tip or hint would be helpful
thanks
Discuss the uniform convergence of the following sequence in the interval indicated
{x^n} , 0< x <1
Now,
f(x) = \lim_{n\rightarrow \infty} f_{n}(x) = 0
Therefore given any small \epsilon > 0 , if there exists N such that |f_n(x)-f(x)| < \epsilon for all n \geq N for all x in the...
I haven't done uniform convergence since last year when I took analysis, and now I have this problem for topology (we're studying metric spaces right now) and I can't remember how to disprove uniform convergence:
f_n: [0,1] -> R , f_n(x)=x^n
Show the sequence f_n(x) converges for all x in...
Suppose (f_n} is a sequence of functions where f_n(x) = x / (1 + n^2 x^2).
I am finding the pointwise limit of the sequence of {f_n'(x)} on the interval
(-oo, + oo)...in which {f_n'(x)} is the sequence of functions obtained from the derivative of x / (1 + n^2 x^2) and I am trying to find...
Hi,
I don't know how to analyse uniform convergence/local uniform convergence for this series of functions:
\sum_{n=1}^{\infty}2^{n}\sin \frac{1}{3^{n}x}
Then
f_{n}^{'} = -\left(\frac{2}{3}\right)^{n}\frac{\cos \frac{1}{3^{n}x}}{x^2}
f_{n}^{'} = 0 \Leftrightarrow x =...
Hi,
I've little probles with this one:
Analyse uniform and local uniform convergence of this series of functions:
\sum_{k = 1}^{\infty} \frac{\cos kx}{k}
I'm trying to solve it using Weierstrass' criterion, ie.
\mbox{Let } f_n \mbox{ are defined on } 0 \neq M \subset...
Hi,
next one I've little problems with:
f_{n} = \frac{2nx}{1+n^{2}x^{2}}
\mbox{a) } x \in [0, 1]
\mbox{b) } x \in (1, \infty)
First the pointwise convergence:
\lim_{n \rightarrow \infty} \frac{2nx}{1+n^{2}x^{2}} = 0
I computed the derivative of my function to...