User:J D Bowen/Math710 HW7

Please do the following problems from Chapter 6 of the text:

10,11,12,13,16,17,20

10) Suppose $f_n \in L^\infty \ $. We aim to show that $f_n\to_{L^\infty} f \iff \exists E: mE=0, \ f_n\to f \ $ uniformly on complement of E.

$\Rightarrow \ $

Suppose $f_n\to f \ $ in $L^\infty \ $, ie, $|f_n-f|\to 0 \ $.

Define $B_{M,n} = \left\{{x:|f_n(x)-f(x)|>M }\right\} \ $.

By definition, $\forall \epsilon \ \exists N: \ n\geq N \implies \text{inf}\left\{{M:mB_{M,n}=0}\right\}<\epsilon \ $.

So $mB_{\epsilon,n}=0 \ $ and $f_n\to f \ $ uniformly on $B_{\epsilon,n}^c \ $, since we have explicitly removed all points where the convergence is not uniform.

$\Leftarrow$

Suppose $\exists E \ $ such that $mE=0 \ $ and $f_n\to f \ $ uniformly on $E^c \ $. Then for all $\epsilon>0, \ \exists N \ $ such that $|f_n(x)-f(x)|<\epsilon \ $ when $n\geq N \ $ for all $x\in E^c \ $. So $B_{\epsilon,n}\subset E \ $, and so $|f_n-f|\to 0 \ $.

11) Let $f_n \ $ be a Cauchy sequence of functions in $L^\infty \ $. Then for all $\epsilon \ $ there exists an $N \ $ such that $n\geq N \implies \text{inf}\left\{{M:mB_{M,n}=0}\right\}<\epsilon \ $, ie, $mB_{M,n}=0 \ $.  In particular, this is true for $M=\epsilon \ $, and therefore $f_n\to f \ $ uniformly on $B_{\epsilon,n}^c \ $.

Now suppose there is a set $E \ $ with $mE =0 \ $ and $f_n\to f \ $ on $E^c \ $. Then there is an $N \ $ such that $|f_n(x)-f(x)|<\epsilon \ $ for $x\in E^c \ $. Therefore, $B_{\epsilon, n}\subset E \ $. But then $\text{inf}\left\{{M:mB_{M,n}=0}\right\}<\epsilon \ $, since $mE=0 \ $, and so $||f_n-f||\to 0 \ $.

12) Define $x_n=\left\{{y_{n1},y_{n2},\dots }\right\} \ $. Suppose $\left\{{x_n}\right\} \ $ is a Cauchy sequence.  Then for all $\epsilon>0 \ $, there is an $N \ $ such that for $m,n\geq N \ $, we have

$||x_m-x_n||_p<\epsilon \ $.

But $||x_m-x_n||_p = \left({ \sum_{i=1}^\infty |y_{mi}-y_{ni}|^p }\right)^{1/p} \ $, so this is just

$\sum_{i=1}^\infty |y_{mi}-y_{ni}|^p < \epsilon^p \ $.

Since $|y_{mi}-y_{ni}|^p>0 \ $, this implies $|y_{mi}-y_{ni}|^p<\epsilon^p \implies |y_{mi}-y_{ni}|<\epsilon \ $, and so there is a sequence $x \ $ such that $x_n\to x \ $.

Observe that for all $\epsilon>0 \ $, there is an $N \ $ such that $n>N\implies ||x_N-x_n||_p < \epsilon \ $. Since $||x_n||_p \ $ is a Cauchy sequence, it converges and so $||x_n||_p \to ||x||_p \ $.

13) Suppose $f_n \in C[0,1] \ $ is a Cauchy sequence. Then $\exists f: \ f_n \to f \ $ pointwise, and for all $\epsilon \  $ there is a $N \ $ such that $n>N\implies ||f_n-f||<\epsilon \ $, so $f_n\to f \ $ uniformly.

Since a uniformly convergent sequence of continuous functions converges to a continuous function, $f \ $ is continuous.

Now define $a_n=\text{max}_{i\leq n} ||f_i|| \ $. Since $f_n\to f \ $, for all $\epsilon>0 \ $ there is an $N \ $ such that $n>N \implies |f_n-f|<\epsilon \ $, so $a_n \ $ is a Cauchy sequence of real numbers and $a_n\to a \ $ satisfies $a-a_n<\epsilon \ $. Therefore, $\text{max}(f) \ $ exists and is the limit of $a_n=\text{max}_{i\leq n} ||f_i|| \ $.

16) Suppose $f_n \to f \ $. Then

$f_n\to f \implies f_n^p\to f^p \implies \int_0^1 |f_n^p|\to\int_0^1 |f^p|\implies \int_0^1 |f_n|^p \to \int_0^1 |f|^p \ $

$\implies \left({ \int_0^1 |f_n|^p }\right)^{1/p} \to \left({ \int_0^1 |f|^p }\right)^{1/p} \implies ||f_n||_p\to ||f||_p \ $.

Now suppose $||f_n-f||_p\to 0 \ $. This is true if and only if $\int \left|{f_n-f}\right| \to 0 \ $.

So let $g_n, f_n, g, f \ $ be as in the theorem and note $|f_n|\leq g_n \implies |f|\leq g \ $. Therefore, $|f-f_n|\leq |f_n|+|f|\leq g_n+g \ $, and so $ h_n=g_n+g-|f_n-f| \geq 0 \ $ is a sequence of nonnegative measurable functions.

By Fatou, we have $\int \lim h_n \leq \lim \inf \int h_n \ $,

which becomes

$\int 2g \leq \int 2g - \lim \sup \int |f_n-f| \ $.

This gives $\lim \sup \int |f_n-f| \leq 0 \leq \lim \inf \int |f_n-f| \implies |f_n-f|\to 0 \ $.

17) Suppose $f_n \to f \ $, where $f_i, f\in L^p \ $, and suppose $||f_k||_p0, \ g\in L^q \ $ and a set $E \ $ there is a $\delta \ $ such that $mE<\delta \implies ||g||_q^q < (\epsilon/4M)^q \ $, since $g \ $ is bounded on $E \ $.

By definition, since $f_n\to f \ $ uniformly on $E^c \ $, there is an $N \ $ such that $x\in E^c, \ n>N \implies |f(x)-f_n(x)|<\epsilon/(2(mE^c)^{1/p}||g||_q) \ $.

Finally, observe that by Holder's inequality,

$\int |fg-f_ng| \leq \int |f-f_n||g| \leq \left({ \int |f-f_n|^p }\right)^{1/p} \left({\int |g|^q}\right)^{1/q} \ $

$=\left({ \int_E |f-f_n|^p }\right)^{1/p} \left({\int_E |g|^q}\right)^{1/q}+\left({ \int_{E^c} |f-f_n|^p }\right)^{1/p} \left({\int_{E^c} |g|^q}\right)^{1/q} \ $

$\leq 2M \epsilon/(4M)+(\epsilon/(2(mE^c)^{1/p}||g||_q))(mE^c)^{1/p}||g||_q = \epsilon \ $.

20) For a partition $\Delta=\left\{{a=\xi_0<\dots<\xi_n=b}\right\} \ $ of $[a,b] \ $ with $\xi_i-\xi_{i-1}\leq \delta \ \forall 1\leq i \leq n \ $ and a function $f:[a,b]\to\mathbb{R} \ $, define

$\varphi_\Delta(x) = \sum_{i=1}^n \left({ \chi_{[\xi_{i-1},\xi_i]}(x)\frac{1}{\xi_i-\xi_{i-1}} \int_{\xi_{i-1}}^{\xi_i} f(t)dt }\right) \ $.

Let $\epsilon>0 \ $ and observe that by the definition of Lebesgue integral, there exists a $\delta $ such that we have

$\int_a^b |f(x)-\varphi_\Delta(x)| dx < \epsilon $,

since $\varphi_\Delta \ $ is a step function sandwiched between the maximal and minimal step functions of $f \ $ on the partition.

Hence for any $\epsilon^{-p} >0 \ $, there is a $\delta \ $ such that $m\left\{{x:|f-\varphi_\Delta|>\epsilon}\right\}<\epsilon \ $. But then by the theorem, we have

$m\left\{{x:|f-\varphi_\Delta|>\epsilon}\right\}=\epsilon m\left\{{x:||f-\varphi_\Delta||_p>\epsilon^{p}}\right\} \ $,

and so $\forall \epsilon, \exists \delta : m\left\{{x:||f-\varphi_\Delta||_p>\epsilon^{p}}\right\}<\epsilon \ $.