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Let $\{f_n\}_{n=1}^{+\infty}$ be a sequence of functions where $f_n(x)=x-{x^n \over n}$ where $x$ belongs to $[0,1]$ and n belongs to $N$ (all natural numbers).

Determine whether the sequences of functions {$f_n$} and {$f'_n$} are uniformly convergent or not.

I'm having trouble with the problem and looking for help. Thanks in advance!

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  • $\begingroup$ To check $f_n$ say, do you know what function it converges to pointwise? Once you determine this, try showing that the supremum of $|f_n(x)-f(x)|$, where $f$ is the limit, goes to $0$ as $n\to\infty$. Hint: Don't be afraid to use calculus. $\endgroup$ – Mike B Apr 11 '13 at 4:05
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The sequence is uniformly Cauchy on $[0,1]$ since $$ |f_n(x) - f_m(x)| = \left|\frac{x^n}{n} - \frac{x^m}{m}\right| \leq \frac{1}{n} + \frac{1}{m} $$ and so this goes to zero in a fashion that does not rely on $x$. Hence, the sequence converges uniformly on $[0,1]$. It must do so to the function $f(x) = x$ since this is what the sequences converges to pointwise. If you want to be more direct about it, note that $$ |f_n(x) - x| = \left|\frac{x^n}{n}\right| \leq \frac{1}{n} $$ for all $x \in [0,1]$ and so the sequence must be uniformly convergent to $f(x) = x$ since the limit does not depend on $x$.

Now the derivative is $f^{\prime}(x) = 1 - x^{n-1}$. This converges pointwise on $[0,1]$ to the piecewise function $$g(x) = \left\{\begin{array}{c@{\qquad}l} 1 & \mbox{if } 0 \leq x < 1 \\ 0 & \mbox{if } x = 1\end{array}\right.$$ A sequences of continuous functions can only converge uniformly to something that is continuous and since $g$ is not continuous, $f^{\prime}$ does not converge uniformly.

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