- #1

twoflower

- 368

- 0

Hi,

next one I've little problems with:

[tex]

f_{n} = \frac{2nx}{1+n^{2}x^{2}}

[/tex]

[tex]

\mbox{a) } x \in [0, 1]

[/tex]

[tex]

\mbox{b) } x \in (1, \infty)

[/tex]

First the pointwise convergence:

[tex]

\lim_{n \rightarrow \infty} \frac{2nx}{1+n^{2}x^{2}} = 0

[/tex]

I computed the derivative of my function to know where to get supremum and I got that maximum is in [itex]x=\frac{1}{n}[/itex].

I used it to get supremum on the first interval and I got

[tex]

\sup \left\{\left|\frac{2nx}{1+n^{2}x^{2}}\right|, x \in [0,1]\right\} = 1 \Rightarrow f_{n} \mbox{ doesn't converge uniformly on } [0,1]

[/tex]

On the second interval [itex](1, \infty)[/itex] I cannot use the point [itex]x = \frac{1}{n}[/itex] and since the function is decreasing on this interval I will take the leftmost point to compute the supremum. It will be

[tex]

\sup \left\{\left|\frac{2nx}{1+n^{2}x^{2}}\right|, x \in (1,\infty)\right\} \leq \left|\frac{2n}{1+n^2}\right| \longrightarrow 0 \mbox{ as n goes to \infty}

[/tex]

Thus

[tex]

f_{n} \rightrightarrows 0 \mbox{ on (1, \infty)}

[/tex]

According to "official" results this is not complete result. Would you help me please to finish it?

Thank you.

next one I've little problems with:

[tex]

f_{n} = \frac{2nx}{1+n^{2}x^{2}}

[/tex]

[tex]

\mbox{a) } x \in [0, 1]

[/tex]

[tex]

\mbox{b) } x \in (1, \infty)

[/tex]

First the pointwise convergence:

[tex]

\lim_{n \rightarrow \infty} \frac{2nx}{1+n^{2}x^{2}} = 0

[/tex]

I computed the derivative of my function to know where to get supremum and I got that maximum is in [itex]x=\frac{1}{n}[/itex].

I used it to get supremum on the first interval and I got

[tex]

\sup \left\{\left|\frac{2nx}{1+n^{2}x^{2}}\right|, x \in [0,1]\right\} = 1 \Rightarrow f_{n} \mbox{ doesn't converge uniformly on } [0,1]

[/tex]

On the second interval [itex](1, \infty)[/itex] I cannot use the point [itex]x = \frac{1}{n}[/itex] and since the function is decreasing on this interval I will take the leftmost point to compute the supremum. It will be

[tex]

\sup \left\{\left|\frac{2nx}{1+n^{2}x^{2}}\right|, x \in (1,\infty)\right\} \leq \left|\frac{2n}{1+n^2}\right| \longrightarrow 0 \mbox{ as n goes to \infty}

[/tex]

Thus

[tex]

f_{n} \rightrightarrows 0 \mbox{ on (1, \infty)}

[/tex]

According to "official" results this is not complete result. Would you help me please to finish it?

Thank you.

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