Uniform Convergence of ##\{f_n\}## on ##[0,a]##

[Lee33]

Homework Statement

Show that the sequence of functions $x,x^2, ...$ converges uniformly on $[0,a]$ for any $a\in(0,1)$, but not on $[0,1]$.2. The attempt at a solution

Is this correct? Should I add more detail? Thanks for your help!

Let $\{f_n\} = \{x^n\}$, and suppose $f^n \to f$. We must show that for $\epsilon>0$, there exists an $N$ such that $d(f,f^n)<\epsilon$ whenever $n>N$ for all $x.$

For $a\in (0,1)$, it is clear to see that $x^n\to 0$ as $n$ approaches infinity. We must then show $|x^n|<\epsilon$ whenever $n$ is greater than some $N$.

On $[0,a]$, $x^n$ attains its max at $x=a$ so $x^n<a^n$. Then note that $a^n$ decreases with increasing $n$, so we choose $N$ such that $a^N<\epsilon$.

$\{f_n\}$ does not converge uniformly on $[0,1]$ because at $x=1$, $f^n = (1)^n =1\ne 0$ for all $n$.

 

[LCKurtz]

Homework Statement

Show that the sequence of functions $x,x^2, ...$ converges uniformly on $[0,a]$ for any $a\in(0,1)$, but not on $[0,1]$.2. The attempt at a solution

Is this correct? Should I add more detail? Thanks for your help!

Let $\{f_n\} = \{x^n\}$, and suppose $f^n \to f$. We must show that for $\epsilon>0$, there exists an $N$ such that $d(f,f^n)<\epsilon$ whenever $n>N$ for all $x.$

For $a\in (0,1)$, it is clear to see that $x^n\to 0$ as $n$ approaches infinity. We must then show $|x^n|<\epsilon$ whenever $n$ is greater than some $N$.

On $[0,a]$, $x^n$ attains its max at $x=a$ so $x^n<a^n$. Then note that $a^n$ decreases with increasing $n$, so we choose $N$ such that $a^N<\epsilon$.

$\{f_n\}$ does not converge uniformly on $[0,1]$ because at $x=1$, $f^n = (1)^n =1\ne 0$ for all $n$.

You need more for the case $x\in [0,1]$. You haven't shown that the convergence of $x^n$ to the function$$
f(x) = \left\{\begin{array}{rl}
0,& 0\le x<1\\
1, & x=1
\end{array}\right.$$is not uniform. After all, that is the function to which it converges for each $x$. Why isn't the convergence uniform?

 

[Lee33]

LCKurtz - Can you elaborate?

For $[0,1]$, $\{f_n\}$ does not converge uniformly on $[0,1]$ because at $x=1$, $f^n = (1)^n =1\ne 0$ for all $n$.

 

[LCKurtz]

LCKurtz - Can you elaborate?

For $[0,1]$, $\{f_n\}$ does not converge uniformly on $[0,1]$ because at $x=1$, $f^n = (1)^n =1\ne 0$ for all $n$.

You know the definition of uniform convergence. Given $\epsilon > 0$ you can find $N$ such that if $n>N$, $\|f_n-f\| < \epsilon$. To show this is false you need to find an $\epsilon$ for which you can't find an $N$ that works. You already know the problem is near $x=1$. Think about that.

 

[Lee33]

Okay, thanks.

 

[LCKurtz]

Okay, thanks.

Please note I edited my post. It's late and the first one needed fixing.

 

[Lee33]

Its okay, thanks for the help!