Question

suppose that for n in Natural

fn(x)={ -1, -1le x le 1 le -1/n
{ nx, |x| le 1/n
{ 1 , 1/n le x le 1

a)find the limit function for {fn}
b)determine whether {fn} converges uniformly on [-1,1]
c)show that int_{-1}^{1} lim_{n rightarrow infty } fn(x) dx =lim_{n rightarrow infty } int_{-1}^{1} fn(x) dx

Answer 1

i don't know what an .nb file is

Answer 2

Mathematica notebook.

Answer 3

hope you got my somewhat sketchy answer to last problem on uniform convergence

Answer 4

\[f_n(x) = \left\{\begin{array}{rcc}
-1 & \text{if} &-1\leq x \leq-\frac{1}{n} \\
nx& \text{if} &-\frac{1}{n}\leq x \leq \frac{1}{n}\\
1 & \text{ if } & \frac{1}{n}\leq x\leq 1
\end{array}
\right.
\]
is my guess

Answer 5

\[f(x) = \left\{\begin{array}{rcc}
-1& \text{if} & x < 0\\
1& \text{if} & x >0
\end{array}
\right.\]

Answer 6

yes for fn(x) as ur guess

Answer 7

ok good
i would imagine the convergence is NOT uniform, for given any N there will be an \(x\) in \((-\frac{1}{N},\frac{1}{N})\) with \(x=\frac{1}{2N}\) and so for that \(x\) \(f_N(x)=\frac{1}{2}\)

Answer 8

as for C
\[\int_{-1}^{1} \lim_{n \to \infty } fn(x) dx=\int_{-1}^1f(x)dx\]the piecewise function above, which is fairly clearly zero. you would integrate
\[\int_{-1}^0-1dx+\int_0^11dx=-1+1=0\]

Answer 9

for the last part, need an expression for
\[\int_{-1}^1f_n(x)dx\] which will require breaking the integral in to three parts, from -1 to \(-\frac{1}{n}\) then from \(-\frac{1}{n}\) to \(\frac{1}{n}\) then from \(\frac{1}{n}\) to 1
the middle one is zero i think, but you can check so it really amounts to
\[\int_{-1}^{-\frac{1}{n}}-1dx+\int_{\frac{1}{n}}^1 1dx\] which by geometry is
\[-1+\frac{1}{n}+1-\frac{1}{n}\]

Answer 10

can you explain little bit about b) determine whether {fn} converges uniformly on [-1,1]. i i cannot see their ways to solve