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 $$f_n\to_{L^\infty} f \iff \exists E: mE=0, \ f_n\to_{\text{uniformly on complement of E}} f \ $$.

$$\Rightarrow \ $$

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