Does the value of p-q affect the uniform convergence of fn on [0,1-ε]?

[cummings12332]

Homework Statement

define function fn:[0,1]->R by fn(x)=(n^p)x*exp(-(n^q)x) where p , q>0 and
fn->0 pointwise on[0,1] as n->infinite.,with the pointwise limit f(x)=0 ,and sup|fn(x)|=(n^(p-q))/e
assume that ε is in (0,1)
does fn converges uniformly on [1,1-ε]? how about on[0.1-ε]?

The Attempt at a Solution

my idea is checking whether the pointwise limit f is continuous on the interval above,but it is obvious,then f is continuous ,so fn is uniformly convergent,but i thought it is wrong. can someone give me any idea?

 

[SammyS]

Homework Statement

define function fn:[0,1]->R by fn(x)=(n^p)x*exp(-(n^q)x) where p , q>0 and
fn->0 pointwise on[0,1] as n->infinite.,with the pointwise limit f(x)=0 ,and sup|fn(x)|=(n^(p-q))/e
assume that ε is in (0,1)
does fn converge uniformly on [1,1-ε]? how about on[0.1-ε]?

The Attempt at a Solution

my idea is checking whether the pointwise limit f is continuous on the interval above,but it is obvious,then f is continuous ,so fn is uniformly convergent,but i thought it is wrong. can someone give me any idea?

First of all, that looks like a typo → does fn converge uniformly ... how about on[0.1-ε]. Shouldn't that be [0,1-ε] ?

As for the problem:

The answer to this question may depend upon the values of p & q, or at least upon the value of p-q .

Consider the supremum: of f_n which is: $\displaystyle\frac{n^{p-q}}{e}\ .$

At what value of x, does the supremum occur?

 

[cummings12332]

First of all, that looks like a typo → does fn converge uniformly ... how about on[0.1-ε]. Shouldn't that be [0,1-ε] ?

As for the problem:

The answer to this question may depend upon the values of p & q, or at least upon the value of p-q .

Consider the supremum: of f_n which is: $\displaystyle\frac{n^{p-q}}{e}\ .$

At what value of x, does the supremum occur?

yes, i know that the supremum occur at 1/(n^q),which is always smaller or equal to 1, but what's the relation for this x value and the interval. i had proved that this function if p<q it is convergent uniformly on [0,1] ,if p>=q it is not uniformly convergent on [0,1] because the supremum goes to infinite. but i cannot see what's the different when i change the intevrval to [0,1-ε]..