Show that the sequence fn(x)=xn converges pointwise but not uniformly on [0,1].
Get the complete, step-by-step math solution for: "Show that the sequence f_n(x) = x^n converges pointwise but not uniformly on [0,1].". Powered by SolveForX AI math tutor.
Step-by-Step Solution
Step 1: Determine Pointwise Limit
To find the pointwise limit of the sequence
fn(x)=xn
on the interval
[0,1]
, we need to evaluate the limit of
xn
as
n
approaches infinity for each
x
in the interval. We consider three cases for
x
:
x=0
,
0<x<1
, and
x=1
.
Step 2: Calculate Pointwise Limit for Cases
For
x=0
,
n→∞lim0n=0
. For
0<x<1
,
n→∞limxn=0
. For
x=1
,
n→∞lim1n=1
. Thus, the pointwise limit function f(x) is
0
for
0≤x<1
and
1
for
x=1
.
Step 3: Check for Uniform Convergence
To check for uniform convergence, we need to evaluate
Mn=x∈[0,1]sup∣fn(x)−f(x)∣
. If
n→∞limMn=0
, then the convergence is uniform. Otherwise, it is not uniform.
Step 4: Calculate
Mn
For
0≤x<1
,
f(x)=0
, so
∣fn(x)−f(x)∣=∣xn−0∣=xn
. For
x=1
,
fn(1)=1n=1
and
f(1)=1
, so
∣fn(1)−f(1)∣=∣1−1∣=0
. Therefore, we need to find the supremum of
xn
on the interval
[0,1)
.
Step 5: Evaluate
n→∞limMn
As
x
approaches
1
from the left,
xn
approaches
1
. Since
xn
is an increasing function on
[0,1)
, the supremum of
xn
on
[0,1)
is
1
. Therefore,
Mn=1
for all
n
.
Step 6: Conclusion
Since
n→∞limMn=1e0
, the sequence
fn(x)=xn
does not converge uniformly on
[0,1]
. However, we have already shown that it converges pointwise to
f(x)={01if 0≤x<1if x=1
.