※ 引述《hcsoso (索索)》之銘言：
: 今天在寫 Rudin 時和同學討論的到的問題，
: 想要請問關於 (c) 小題的解題方向...
: 那一小題主要是在問
: 2xf(x) = cos(x^2) - cos((x+1)^2) + r(x) where |r(x)| < c/x 時，
: limsup f(x) 和 liminf f(x) 分別是多少，當 x -> infinity
: 我們可以發現 limsup f(x) 應該要是 1，並且也已經可以證明 <= 1，
: 但對於另一個方向實在沒有頭緒...
: 想試著證對於所有 epsilon 以及對於所有 M，
: 我們都有 x >= M 使得 f(x) > 1 - epsilon，
: 可是卻做不出來...
: 想請問板上的大大們，這樣的方向正確嗎？
: 如果不正確，那應該朝什麼方向嘗試呢？
: 謝謝！
(c) Find the upper and lower bound limit of xf(x), as x→∞.
Proof. [這是我好幾年前寫的，有問題再討論討論吧 XD]
We will use Lemma (in note) to show lim sup xf(x)＝1, and lim inf xf(x)＝－1.
Claim lim sup cos (x^2)－cos [(x＋1)^2]＝2 as follows. Taking x ＝ n√(2π),
where n in Z, then
cos (x^2)－cos [(x＋1)^2] ＝ 1－ cos (n√(8π) ＋ 1)
If we can show that {n√(8π)} is dense in [0,2π) mod 2π, we complete the
proof. It is equivalent to show that {n√(2/π)} is dense in [0,1) mod 1.
It is clear that by the lemma in the note. So, we have proved that
lim sup cos (x^2)－cos [(x＋1)^2]＝2. Similarly, we also have
lim inf cos (x^2)－cos [(x＋1)^2]＝－2.
So, we finally have lim sup xf(x)＝1, and lim inf xf(x)＝－1.
Note. (Lemma－Dirichlet) If α is an irrational number, then
S ＝ {nα＋m: n in N, m in Z} is dense in |R. Equivalently;
{nα} is dense in [0,1) modulus 1.
[The proof of lemma can be found in the Exercise 25, Chap 3.]
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