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Pick out the sequences $\{f_n\}$ which are uniformly convergent.
(a) $f_n(x) = nxe^{−nx}$ on $(0,∞)$.

(b)$f_n(x) = x^n$ on $[0, 1]$.

(c)$f_n(x) = \frac{\sin(nx)}{\sqrt{n}}$ on $\mathbb{R}$.

(d) $f_n(x)=\frac{nx}{1 + nx}$ on $(0,1)$

(e) $f_n(x) = ∑_{n=1}^∞\frac{n\sin(nx)}{e^n}$ on $[0,\pi]$

(f) $f_n(x) = \frac{x^n}{1 + x^n}$ ; on $[0, 2]$

(g) $f_n(x) = \sin^nx$ on $[0,\pi/2)$

(h) $f_n(x) = (x^n/n)+1$ on $[0,1)$

(i) $f_n(x) = \frac{1}{1+(x-n)^2}$ on $(-∞,0)$

(j) $f_n(x) = \frac{1}{1+(x-n)^2}$ on $(0, ∞)$

(a) Here the function becomes $nx/e^{nx}$ which is tends to $0$ as $n$ tends to $∞$ so the function is uniformly continuous. I am not sure though.
(b) Here limit function is not continuous so not uniformly continuous
(c) true by same reason of (a) (d) false as $n$ tends to $∞$ the function tends to $1$ not zero.
(e) no idea (f) false as limit function is not continuous (g) true by M test (h) true by M test (i) no idea (j) no idea

Can somebody guide me properly please

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Woooo! Many questions here, poton. – Nancy Rutkowskie Dec 31 '12 at 16:17
Do you know the definition of uniformly convergent ? – Siméon Dec 31 '12 at 16:21
up vote 3 down vote accepted

A related technique. Note that, for (e), you can use the Weierstrass M-Test, since

$$\left| \frac{n\sin(nx)}{e^n} \right| \leq \left| \frac{n}{e^n} \right|=\frac{n}{e^n}, $$


$$ \sum_{n=1}^{\infty}\frac{n}{e^n} < \infty. $$

Note: if

$$\lim_{ n \to \infty}f_n(x)=f(x) \quad\mathrm{ and } \quad \sup| f_n(x)-f(x) | < \epsilon, $$

then the sequence converges uniformly.

For (j), notice that, $f_n(x)\to 0$ for $x\neq n$ and $f_n(x)\to 1$ for $x = n$ which implies

$$ \sup |f_n(x)-f(x)| =1 \nless \epsilon. $$

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The general idea is to fix every $x$,and let $n$ goes to infinity. Then you get a function $f$ which is the limit of $f_n$ point-wise. Afterwards you can use the Weierstrass-M test for $f_n$.

For (a), the point-wise limit function is $0$, and you calculate the maximal value of each $f_n$, and you will find out that the maximal values goes to $0$, so for any $\varepsilon$, you may get a certain $N$, such that when $k>N$, $f<\varepsilon$ for any $x$ on $(0,\infty)$.

For (b), you should know that any limit of uniformly convergent functions are continuous if $f_n$ are continuous.

(c) is the same approach as (a), and now you know that $\sin(t)$ is bounded for any t.

(d) is not because for every fixed $x$, the limit function you get should be 1, however, for any given n, you may pick $x=\frac{1}{n}$,so that $f_n(x)=\frac{1}{2}$, contradiction.

For (e), you will need the Abel's Criterion, which deals with $\sum_{1}^{\infty}{f_n \times g_n}$, where $f_n$ is decreasing and converges to $0$ and $\sum g_n$ is bounded. Then you can say $\{f_ng_n\}$ is uniformly convergent.

(f) (g) (h) is just like you said, and the approach is similar to the solutions above.

For (i) you should also use the approach in (a), and you find out it's uniformly convergent.

For(j) you pick $x_n=n$ for every given $n$, and you find out it's 1, not converging to the point-wise limit function $0$.

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You need to review your definition of a uniform convergence : $f_n$ converges uniformly to $f$ if for every $\epsilon>0$ there is an integer $N$ such that if $n\ge N$ we have $|f_n-f|<\epsilon$

The Cauchy criterion for uniform convergence is the following $fn$ converges to some function $f$ uniformly if and only if for every $\epsilon>0$ there is an integer $N$ such that if $m>n\ge N$ we have $|f_m-f_n|<\epsilon$ . In (b) don't you mean since the limit function is not continuous then the sequence does not converge uniformly

In (d) that does not imply the sequence is not uniformly convergent it shows the limit function is $1$ , you now need to use the first definition to see whether it is uniformly convergent to 1 or not.

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