1994 Nov

Instructions. Masters students: Do any 5 problems. Ph.D. students: Do any 6 problems.

1. Let \(E\) be a normed linear space. Show that \(E\) is complete if and only if, whenever \(\sum_1^\infty \|x_n \| < \infty\), then \(\sum_1^\infty x_n\) converges to an \(s\in E\).

2. Let \(f_n\) be a sequence of real continuous functions on a compact Hausdorff space \(X\). Show that if \(f_1 \geq f_2 \geq f_3 \geq \cdots\), and \(f_n(x) \rightarrow 0\) for all \(x\in X\), then \(f_n \rightarrow 0\) uniformly.

3. Let \(f\) be integrable on the real line with respect to Lebesgue measure. Evaluate \(\lim_{n\rightarrow \infty} \int_{-\infty}^\infty f(x-n) \left(\frac{x}{1+|x|}\right)\, dx.\) Justify all steps.

Solutions

Solution 1

Suppose \(E\) is complete.

Let \(\{xₙ\} ⊂ E\) be absolutely convergent; that is, \(∑ ∥ xₙ ∥ < ∞\).

We must show, \(∑ₙ^∞ xₙ := \lim_{N → ∞} ∑_{n=1}^N xₙ = s\), for some \(s ∈ E\).

Let \(S_N = ∑_{n=1}^N xₙ\).

For each \(j ∈ ℕ\),

\[ ∥ S_{N+j} - S_N ∥ = \left\|\sum_{n=N+1}^{N+j}x_n\right\| \leq \sum_{n=N+1}^{N+j}\|x_n\| \rightarrow 0 \]

as \(N → ∞\), since \(∑ ∥ xₙ ∥ < ∞\).

Therefore, \(\{S_N\}\) is a Cauchy sequence.

Since \(E\) is complete, \((∃s ∈ E) (∑_{n=1}^∞ xₙ = \lim\limits_{N → ∞} S_N = s)\).

Conversely, suppose the series \(∑₁^∞ xₙ\) converges to an \(s ∈ E\) whenever \(∑₁^∞ ∥ xₙ ∥ < ∞\).

Let \(\{yₙ\} ⊂ E\) be a Cauchy sequence; that is \(∥ yₙ - yₘ ∥ → 0\) as \(n, m → ∞\).

Let \(n₁ < n₂ < ⋯\) be a subsequence such that \(∥ yₙ - yₘ ∥ < 2^{-j}\) for all \(n, m ≥ nⱼ\).

Observe that, for \(k > 1\), we have

\(y_{nₖ} = y_{n₁} + (y_{n₂} - y_{n₁}) + (y_{n₃} - y_{n₂}) + ⋯ + (y_{nₖ} - y_{n_{k-1}}) = y_{n₁} + ∑_{j=1}^{k-1} (y_{n_{j+1}} - y_{nⱼ})\),

and \(∑_{j=1}^∞ ∥ y_{n_{j+1}} - y_{n_j} ∥ < ∑_{j=1}^∞ 2^{-j} = 1\).

By hypothesis, this implies

\(y_{n_k} - y_{n_1} = ∑_{j=1}^{k-1} (y_{n_{j+1}} - y_{n_j}) → s ∈ E\), as \(k → ∞\).

We have thus found a subsequence \(\{y_{n_k}\} ⊆ \{y_n\}\) with a limit in \(E\).
Finally, since \(\{yₙ\}\) is Cauchy, it is easy to check that \(\{yₙ\}\) converges to the same limit.

This proves that every Cauchy sequence in \(E\) converges to a point in \(E\). ∎

Solution 3

Fix \(n>0\).

Consider the change of variables, \(y = x-n\).

Then \(dy = dx\) and \(x = y+n\), so

\(∫_{-∞}^∞ f(x-n) \frac{x}{1+|x|} \, dx = ∫_{-∞}^∞ f(y) \frac{y+n}{1+|y+n|}\, dy =∫_{\{y+n ≥ 0\}} f(y) \frac{y+n}{1+y+n}\, dy +\int_{\{y+n < 0\}} f(y) \left(\frac{y+n}{1-(y+n)}\right)\, dy\)

and

\(∫_{-∞}^∞ f(x-n) \frac{x}{1+|x|}\, dx = ∫_{-∞}^∞ f(y) \frac{y+n}{1+|y+n|}\, dy =∫_{-n}^∞ f(y) \frac{y+n}{1+y+n}\, dy + ∫_{-∞}^{-n} f(y) \frac{y+n}{1-(y+n)}\, dy\).

Note that, when \(y ≥ -n\), \(\frac{y+n}{1+y+n} \in [0,1)\), and increases to \(1\) as \(n\) tends to infinity.

Thus, \(0 ≤ |f(y)|\, \frac{y+n}{1+y+n} ≤ |f(y)|\), for all \(y ≥ -n\).

Define the function \(gₙ(y) = f(y)\, \frac{y+n}{1+y+n} \, 𝟙_{[-n,∞)}(y)\).

(Here \(𝟙_A(y)\) denotes the indicator function of the set \(A\), which is 1 if \(x ∈ A\) and 0 if \(x ∉ A\).)

Then \(|gₙ| ≤ |f|\) and \(\lim\limits_{n → ∞} gₙ = f\). Therefore, by the dominated convergence theorem, \(\lim_{n → ∞} ∫_{-n}^∞ f(y) \frac{y+n}{1+y+n}\, dy = \lim_{n → ∞}∫_{-∞}^∞ gₙ(y) \, dy =∫_{-∞}^{∞} f(y) \, dy\).

Define the function \(hₙ(y) = f(y)\, \frac{y+n}{1-(y+n)} \, 𝟙_{(-∞, -n]}(y)\).

It is not hard to check that \(\frac{|y+n|}{|1-(y+n)|} \, 𝟙_{(-∞, -n]}(y) ∈ [0,1)\),

from which it follows that \(|hₙ| ≤ |f|\).

Also, for all \(y\), \(\lim_{n → ∞} hₙ(y) = f(y)\, \lim_{n → ∞} \frac{y+n}{1-(y+n)} \, 𝟙_{(-∞, -n]}(y) = 0\).

Therefore, the dominated convergence theorem implies \(\lim_{n → ∞} ∫_{-∞}^{-n} f(y) \frac{y+n}{1-(y+n)}\, dy = 0\).

Combining the two results above, we see \(\lim\limits_{n → ∞} ∫_{-∞}^∞ f(x-n) \left(\frac{x}{1+|x|}\right)\, dx = ∫ f(x) \, dx\). ∎

Remark. Intuitively, this is the result we expect because the translation \(f(x-n) = Tₙ f(x)\) is merely shifting the support of \(f\) to the right tail of the measure \(dμ = \frac{x}{1+|x|}\, dx\), which approaches \(dx\) as \(x → ∞\).