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--- math104-s21:s:antonthan [2021/05/07 09:30]
45.48.153.5 [Problem 15]
+++ math104-s21:s:antonthan [2026/02/21 14:41] (current)
@@ Line 50: / Line 50: @@
 === (a) ===
-We show that Cauchy condition on $B(X)$ implies that the sequence converges (uniformly) to a function $f(x)$, which can be found through pointwise convergence.  We can then use the uniform convergence to show that $f(x)$ is bounded (and obviously, defined on every point in $X$), so it is an element of $B(X)$.  So, every Cauchy sequence converges in $B(X)$, so the metric space is complete.
+We show that Cauchy condition on $B(X)$ is the definition of being uniformly Cauchy.  Uniformly Cauchy implies uniform convergence, and our convergent function is bounded, so $B(X)$ is complete.
 === (b) ===
@@ Line 57: / Line 57: @@
 Therefore, the set of continuous functions is closed.
+Easier way: consider that a set is closed iff it contains all its limit points, and consider a sequence of continuous functions that uniformly converges.
 === (c) ===
@@ Line 145: / Line 147: @@
 ==== Problem 14 ====
+We consider the function $g(x) = f(x) - f(x + T/2)$, which is continuous.  Show that $g(0) = -g(T/2)$, and then use IVT.
 ==== Problem 15 ====
@@ Line 163: / Line 165: @@
 ==== Problem 17 ====
+Consider $f(x)e^{g(x)}$, which is continuous and differentiable on the same intervals.  We take its derivative and apply MVT, and we are done.
 ==== Problem 18 ====
 === (a) ===
+Let partition $P_n = \{0, 1/n, 2/n, \ldots, n/n\}$.  Note that $mesh(P_n) = 1/n$ and $L(f, P_n) \leq R_n \leq U(f, P_n)$.
+The idea now is to use Ross 32.7.  Let $\epsilon > 0$, and since $f$ is integrable, there exists a $\delta > 0$ that satisfies $mesh(P) < \delta \implies U(f, P) - L(f, P) < \epsilon$ for all partitions $P$.  We choose $n$ large enough such that $mesh(P_n) < \delta$, so that we get the inequalities: $U(f, P_n) - \epsilon < R_n \leq U(f, P_n$ and $L(f, P_n) \leq R_n < L(f, P_n) + \epsilon$.  This gives us:
+$$U(f) \leq \sup_n(U(f, P_n)) = lim R_n = \inf_n(L(f, P_n)) \leq L(f)$$
+Since we also have $L(f) = U(f)$ by integrability, $lim R_n = \int_0^1 f(x)dx$, as desired.
 === (b) ===
@@ Line 385: / Line 396: @@
 ===== ඞ ඞ ඞ ඞ ඞ =====
-{{ :math104:s:amongus.mp4 |}}
+{{ math104-s21:s:amongus.mp4 |}}

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