User:MCPOliseno /Math710 HW 7

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, $$|f_n-f|\to 0 \ $$. Then define $$B_{M,n} = \left\{{x:|f_n(x)-f(x)|>M }\right\} \ $$. Then by definition, $$\forall \epsilon \ \exists N: \ n\geq N \implies \text{inf}\left\{{M:mB_{M,n}=0}\right\}<\epsilon \ $$. Thus $$mB_{\epsilon,n}=0 \ $$ 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 $$ \forall \epsilon>0 \exists N \ $$ such that $$|f_n(x)-f(x)|<\epsilon \ $$ when $$n\geq N \ $$ $$ \forall x\in E^c \ $$. Thus $$B_{\epsilon,n}\subset E \ $$, and therefore $$|f_n-f|\to 0 \ $$.

12 Define $$x_n=\left\{{y_{n1},y_{n2},\dots }\right\} \ $$. Suppose $$\left\{{x_n}\right\} \ $$ is a Cauchy sequence. Then $$ \forall \epsilon>0 \ $$, $$ \exists \ $$ an $$N \ $$ such that for $$m,n\geq N \ $$, we have $$||x_m-x_n||_p<\epsilon \ $$. But $$||x_m-x_n||_p = \left({ \sum_{i=1}^\infty |y_{mi}-y_{ni}|^p }\right)^{1/p} \ $$, thus $$\sum_{i=1}^\infty |y_{mi}-y_{ni}|^p < \epsilon^p \ $$. Then, 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 $$x_n\to x \ $$. Note $$ \forall \epsilon>0 \exists \ $$ an $$N \ $$ such that $$n>N\implies ||x_N-x_n||_p < \epsilon \ $$. Thus, since $$||x_n||_p \ $$ is a Cauchy sequence, it converges and $$||x_n||_p \to ||x||_p \ $$.

13 Suppose $$f_n \in C[0,1] \ $$ is a Cauchy sequence. Then $$\exists f: \ f_n \to f \ $$ pointwise, and for all $$\epsilon \exists \ $$  $$N \ $$ such that $$n>N\implies ||f_n-f||<\epsilon \ $$, thus $$f_n \to f \ $$ uniformly. Then, 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 Assume that $$f_n \to f \ $$. 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 lets suppose $$||f_n-f||_p\to 0 \ $$, if and only if $$\int \left|{f_n-f}\right| \to 0 \ $$. Now, 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 seq. of nonnegative measurable functions. Then by Fatou's lemma, we have $$\int \lim h_n \leq \lim \inf \int h_n \ $$, which then becomes $$\int 2g \leq \int 2g - \lim \sup \int |f_n-f| \ $$. Thus, we get $$\lim \sup \int |f_n-f| \leq 0 \leq \lim \inf \int |f_n-f| \implies |f_n-f|\to 0 \ $$.

17 Assume $$f_n \to f \ $$, where $$f_i, f\in L^p \ $$, and suppose $$||f_k||_p0, \ $$ \ $$ g\in L^q \ $$ and set $$E \exists \ $$ $$\delta \ $$ such that $$mE<\delta \implies ||g||_q^q < (\epsilon/4M)^q \ $$, since $$g \ $$ is bounded on $$E \ $$. Since $$f_n\to f \ $$ uniformly on $$E^c \ $$, $$ \exists \ $$ an $$N \ $$ such that $$x\in E^c, \ n>N \implies |f(x)-f_n(x)|<\epsilon/(2(mE^c)^{1/p}||g||_q) \ $$. Then 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 Fix $$y \in \mathbb{R} \ $$ and $$f:A \to A \ $$, where $$A \ $$ is a bounded subset of $$\mathbb{R} \ $$. Now define $$E=\left\{{x :f(x)>y}\right\} \ $$. Use partition $$\Delta=\left\{{\xi_0,\dots,\xi_n}\right\}, \ \xi_j-\xi_{j-1}\leq \delta \ \forall j \ $$ of $$A \ $$, define $$F_\Delta = \left\{{x: \varphi_\Delta(x)>y }\right\} \ $$. Then suppose $$(\xi_j,\xi_{j+1})\subset E \ $$. Then $$f(x)>y \ \forall x \in (\xi_j,\xi_{j+1}), \ $$ and thus $$\varphi_\Delta(x)=\frac{1}{\xi_{j+1}-\xi_j}\int_{\xi_j}^{\xi_{j+1}} f(t)dt > \frac{1}{\xi_{j+1}-\xi_j}\int_{\xi_j}^{\xi_{j+1}} ydt=y \ $$. Therefore $$(\xi_j,\xi_{j+1})\subset E \implies (\xi_j,\xi_{j+1})\subset F_\Delta \ $$. Now suppose $$(\xi_j,\xi_{j+1})\subset E^c \ $$. Then $$f(x)<y \ \forall x \in (\xi_j,\xi_{j+1}), \ $$ and thus $$\varphi_\Delta(x)=\frac{1}{\xi_{j+1}-\xi_j}\int_{\xi_j}^{\xi_{j+1}} f(t)dt < \frac{1}{\xi_{j+1}-\xi_j}\int_{\xi_j}^{\xi_{j+1}} ydt=y \ $$. Therefore $$(\xi_j,\xi_{j+1})\subset E^c \implies (\xi_j,\xi_{j+1})\subset F_\Delta^c \ $$. Therefore, the symmetric difference $$E\ominus F_\Delta\subseteq \Delta\cup\bigcup (\xi_k,\xi_{k+1}) \ $$, where the union is taken over all $$(\xi_k, \xi_{k+1}) \ $$ such that $$\exists x_1, x_2 \in (\xi_k, \xi_{k+1}): \ f(x_1)\geq y, f(x_2)\leq y \ $$. Then, since $$\Delta \ $$ is a finite number of points, it has measure zero, and therefore $$m(E\ominus F_\Delta) \leq \Sigma (\xi_{k+1}-\xi_k) \leq m\delta $$, where $$m \ $$ is the number of intervals such that $$\exists x_1, x_2 \in (\xi_k, \xi_{k+1}): \ f(x_1)\geq y, f(x_2)\leq y \ $$. Note that since $$f\in L^P(A) \ $$, it is either the constant $$y \ $$ or itself.