Uniform Convergence of sequence

[siddharth]

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 given interval, then f_n(x) is uniformly convergent.

That gives

$$x^n < \epsilon$$

$$n > \frac{\log \epsilon }{\log x}$$

So, it is not possible to fix an $$N$$ such that the above condition is satisfied for all values of n>N because for a given value of N, I can always find a value of x close to 1 such that the above condition is not valid.
Hence $$x^n$$ is not uniformly convergent in the given interval.

Is my above reasoning correct?

 

[AKG]

xⁿ is convergent, just not uniformly convergent. Does that make sense?

 

[siddharth]

Yes, it does make sense. The reason I asked this is because I was confused after reading this thread.
https://www.physicsforums.com/showthread.php?t=92619"