Worked examples · Sheet 2

Uniform convergence with the sup criterion

Theory used: Uniform convergence Sup criterion

Exercise 1

Study the pointwise and uniform convergence on $[0, + \infty)$ of the sequence of functions

f_{n} (x) = x e^{- 2 n x} .

f_{n} (0) = 0, x > 0 : e^{- 2 n x} \to 0 ⟹ f (x) = 0 on [0, + \infty)

f_{n}^{'} (x) = e^{- 2 n x} (1 - 2 n x) f_{n}^{'} (x) > 0 ⟺ x < \frac{1}{2 n}

M_{n} = x \geq 0 sup ∣ f_{n} (x) - 0∣ = f_{n} (\frac{1}{2 n}) = \frac{1}{2 n e} n \to + \infty 0

f_{n} \to 0 uniformly on [0, + \infty)

The pointwise limit is computed for fixed $x$ : at $x = 0$ every $f_{n}$ equals $0$ , and for $x > 0$ the exponential $e^{- 2 n x}$ tends to $0$ faster than any constant, so $f (x) = 0$ at all points. By the sup criterion we study $∣ f_{n} - f ∣ = f_{n}$ , which is nonnegative. The derivative $f_{n}^{'} (x) = e^{- 2 n x} (1 - 2 n x)$ vanishes at $x = \frac{1}{2 n}$ , is positive before and negative after, so that point is the absolute maximum on $[0, + \infty)$ . The maximum value is $M_{n} = \frac{1}{2 n} e^{- 1} = \frac{1}{2 n e}$ . Since $M_{n} \to 0$ , the convergence is uniform on the whole half-line.

Exercise 2

Study the pointwise and uniform convergence on $R$ of the sequence of functions

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

f_{n} (0) = 0, x \neq = 0 : ∣ f_{n} (x) ∣ \leq \frac{∣ x ∣}{n ^{2} x ^{2}} = \frac{1}{n ^{2} ∣ x ∣} \to 0 ⟹ f (x) = 0 on R

f_{n}^{'} (x) = \frac{1 - n ^{2} x ^{2}}{( 1 + n ^{2} x ^{2} ) ^{2}} f_{n}^{'} (x) = 0 ⟺ x = \pm \frac{1}{n}

M_{n} = x \in R sup ∣ f_{n} (x) ∣ = f_{n} (\frac{1}{n}) = \frac{\frac{1}{n}}{2} = \frac{1}{2 n} n \to + \infty 0

f_{n} \to 0 uniformly on R

For the pointwise limit we fix $x$ : at $x = 0$ all the $f_{n}$ equal $0$ , and for $x \neq = 0$ the denominator grows like $n^{2}$ , so the absolute value of $f_{n} (x)$ is bounded by $\frac{1}{n ^{2} ∣ x ∣}$ , which tends to $0$ . The function $f_{n}$ is odd, therefore the sup of the absolute value on $R$ coincides with the maximum on $[0, + \infty)$ . The derivative vanishes at $x = \frac{1}{n}$ , with $f_{n}$ increasing before and decreasing after, so the maximum equals $f_{n} (\frac{1}{n}) = \frac{1}{2 n}$ . Since $M_{n} = \frac{1}{2 n} \to 0$ , the sup criterion guarantees uniform convergence on all of $R$ .

Exercise 3

Study the pointwise and uniform convergence on $R$ of the sequence of functions

f_{n} (x) = n x e^{- n x^{2}} .

f_{n} (0) = 0, x \neq = 0 : n e^{- n x^{2}} \to 0 ⟹ f (x) = 0 on R

f_{n}^{'} (x) = n e^{- n x^{2}} (1 - 2 n x^{2}) f_{n}^{'} (x) = 0 ⟺ x = \pm \frac{1}{2 n}

M_{n} = x \in R sup ∣ f_{n} (x) ∣ = f_{n} (\frac{1}{2 n}) = \frac{n}{2 n} e^{- \frac{1}{2}} = \frac{n}{2 e} n \to + \infty + \infty

f_{n} \to 0 pointwise but not uniformly on R

y = x e^{-x x}

y = 4 x e^{-4 x x}

y = 16 x e^{-16 x x}

y = n x e^{- n x^{2}}

for

n = 1, 4, 16

: the maximum, at

x = \frac{1}{2 n}

, equals

\frac{n}{2 e}

and grows without bound

For fixed $x \neq = 0$ the quantity $n e^{- n x^{2}}$ tends to $0$ , since the exponential $e^{- n x^{2}}$ decreases faster than the factor $n$ grows; at $x = 0$ every term equals $0$ . Hence the pointwise limit is the zero function on all of $R$ . For the sup we use the oddness of $f_{n}$ and study the positive half-line: the derivative vanishes at $x = \frac{1}{2 n}$ , which is the maximum point, and the maximum value is $\frac{n}{2 e}$ . This sequence tends to $+ \infty$ , so the sup criterion is not satisfied and the convergence is not uniform on $R$ : near the origin the graphs develop an ever taller peak, which pointwise convergence does not see.