Sample Midterm Solutions Math 127B. Winter, 2005

1. Consider the sequence (f_{n}) of functions f_{n} : ℝ → ℝ defined by

f_{n}(x) = \frac{nx}{\sqrt{1 + n^{2}x^{2}}} .

Find the pointwise limit of this sequence as n → ∞. Does the sequence converge uniformly on ℝ? Justify your answer.

Solution.

• For x > 0,

f_{n}(x) = \frac{1}{\sqrt{1/(n^{2}x^{2}) + 1}} → 1 as n → ∞.

For x < 0,

f_{n}(x) = \frac{−1}{\sqrt{1/(n^{2}x^{2}) + 1}} → −1 as n → ∞.

For x = 0,

f_{n}(0) = 0 → 0 as n → ∞.

Hence f_{n} → f pointwise as n → ∞, where

f(x) = \{ \begin{matrix}1 & for x > 0, \\ 0 & for x = 0, \\ −1 & for x < 0. \end{matrix}

• The sequence cannot converge uniformly because the f_{n} are continuous, f is discontinuous, and the uniform limit of continuous functions is continuous.

2. Let

f_{n}(x) = \frac{nx + sin(nx^{2})}{n}.

Prove that the following limit exists, and compute its value:

lim_{n→∞} ∫_{0}^{1} f_{n}(x) dx.

Solution.

• The sequence (f_{n}) of continuous functions converges uniformly to the the function x on [0, 1]. To prove this, suppose ϵ > 0, and choose N = 1/ϵ. Then if n > N, we have that

|f_{n}(x) − x| = |\frac{sin(nx^{2})}{n}| ≤ \frac{1}{n} < ϵ for all x ∈ [0, 1].

If (f_{n}) is a sequence of continuous functions and f_{n} → f uniformly on [a, b], then

∫_{a}^{b} f_{n}(x) dx → ∫_{a}^{b} f(x) dx.

Therefore for the sequence given in the problem

∫_{0}^{1} f_{n}(x) dx → ∫_{0}^{1} x dx = \frac{1}{2}.

3. Prove that the following series

f(x) = ∑_{n=1}^{∞} \frac{n^{2} + x^{4}}{n^{4} + x^{2}}

converges to a continuous function f : ℝ → ℝ.

Solution.

• Suppose R > 0. Then for all x ∈ [−R, R] we have

|\frac{n^{2} + x^{4}}{n^{4} + x^{2}} | ≤ \frac{n^{2} + x^{4}}{n^{4}}
≤ \frac{1}{n^{2}} + \frac{R^{4}}{n^{4}} .

The series

∑_{n=1}^{∞} \{\frac{1}{n^{2}} + \frac{R^{4}}{n^{4}} \} = ∑_{n=1}^{∞} \frac{1}{n^{2}} + R^{4} ∑_{n=1}^{∞} \frac{1}{n^{4}}

converges, so the Weierstrass M-test implies that the series for f converges uniformly on the bounded interval [−R, R]. The terms in the series are continuous and the uniform limit of continuous functions is continuous, so f is continuous on [−R, R] for every R > 0. Since every x ∈ ℝ lies in such an interval for sufficiently large R, it follows that f is continuous on ℝ.

• Note that the series does not converge uniformly on ℝ, so we can’t use the argument that the sum is continuous on ℝ because the series converges uniformly on ℝ.

4. Determine the radius of convergence R of the power series

f(x) = ∑_{n=0}^{∞} \frac{(−1)^{n}}{2n + 1}x^{2n+1}

= x − \frac{x^{3}}{3} + \frac{x^{5}}{5} − \frac{x^{7}}{7} + ….

Where does the series converge? Prove that

f′(x) = \frac{1}{1 + x^{2}} in |x| < R.

Solution.

• By the root test, and the standard limit n^{1/n} → 1 as n → ∞, the radius of convergence is

R = lim sup_{n→∞} (\frac{1}{2n + 1})^{1/(2n+1)} = 1.

(Alternatively, the ratio test gives the same result.) Hence the series converges in |x| < 1 and diverges in |x| > 1. At x = 1, the series is

1 − \frac{1}{3} + \frac{1}{5} − \frac{1}{7} + …,

which converges by the alternating series test. If x = −1, the series is

− (1 + \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + …) ,

which diverges by comparison with the divergent harmonic series, since

1 + \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + … > \frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \frac{1}{8} + …
> \frac{1}{2} (1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + …)

Thus, the series converges for −1 < x ≤ 1.

• A power series is differentiable inside its interval of convergence and can be differentiated term-by-term. Hence, in |x| < 1 we have

f′(x) = ∑_{n=0}^{∞} (−1)^{n}x^{2n}

= ∑_{n=0}^{∞}(−x^{2})^{n}
= \frac{1}{1 + x^{2}} ,

where we have used the standard sum of a geometric series,

∑_{n=0}^{∞} a^{n} = \frac{1}{1 − a} |a| < 1,

with a = −x^{2}.

5. Suppose that (f_{n}) is a sequence of functions f_{n} : [−1, 1] → ℝ that converges uniformly on [−1, 1] to a function f : ℝ → ℝ. If the limit

lim_{x→0} f_{n}(x) = a_{n}

exists for each n ∈ ℕ, and the limit

lim_{n→∞} a_{n} = a

exists, prove that lim_{x→0} f(x) exists, and

lim_{x→0} f(x) = a

Give a counter-example to show that this result need not be true if (f_{n}) converges to f pointwise, but not uniformly.

Solution.

• Let ϵ > 0 be given. From the uniform convergence, we can choose N_{1} such that n > N_{1} implies that

|f_{n}(x) − f(x)| < \frac{ϵ}{3} for all x ∈ [−1, 1].

From the existence of the limit of (a_{n}), we can choose N_{2} such that n > N_{2} implies that

|a_{n} − a| < \frac{ϵ}{3}.

Choose some N > max{N_{1}, N_{2}}. The existence of the limit of f_{N}(x) as x → 0 implies that there exists δ > 0 such that

|f_{N}(x) − a_{N}| < \frac{ϵ}{3} when 0 < |x| < δ.

It follows that if 0 < |x| < δ, then

|f(x) − a| ≤ |f(x) − f_{N}(x)| + |f_{N}(x) − a_{N}| + |a_{N} − a|
< \frac{ϵ}{3} + \frac{ϵ}{3} + \frac{ϵ}{3}
< ϵ,

which proves that f(x) → a as x → 0.

• Consider the sequence of functions on [−1, 1] defined by

f_{n}(x) = (1 − x^{2})^{n}.

Then f_{n} → f pointwise as n → ∞, where

f(x) = \{ \begin{matrix}0 & for x ≠ 0, \\ 1 & for x = 0. \end{matrix}

By the continuity of f_{n}, we have

a_{n} = lim_{x→0} f_{n}(x) = 1

for every n, and a_{n} → 1 as n → ∞. On the other hand,

lim_{x→0} f(x) = 0 ≠ 1.

• This result says that uniform convergence allows us to exchange the order of the limits:

lim_{x→0} lim_{n→∞} f_{n}(x) = lim_{n→∞} lim_{x→0} f_{n}(x)