Worked examples · Sheet 3

Uniform convergence on subintervals

Theory used: Uniform convergence Sup criterion Continuity of the uniform limit

Exercise 1

Study the pointwise and uniform convergence of $f_{n} (x) = x^{n}$ on $[0, 1]$ , on $[0, 1)$ and on $[0, a]$ with $0 < a < 1$ .

f (x) = {01 if 0 \leq x < 1 if x = 1

[0, 1] : f discontinuous at 1 ⟹ convergence not uniform

[0, 1) : M_{n} = 0 \leq x < 1 sup x^{n} = 1 \neq \to 0 ⟹ convergence not uniform

[0, a] : M_{n} = 0 \leq x \leq a sup x^{n} = a^{n} n \to + \infty 0 ⟹ uniform convergence

The pointwise limit distinguishes two cases: for $0 \leq x < 1$ the geometric sequence $x^{n}$ tends to $0$ , while at $x = 1$ every term equals $1$ . On $[0, 1]$ the $f_{n}$ are continuous and the pointwise limit is discontinuous, so by the theorem on the continuity of the uniform limit the convergence cannot be uniform. On $[0, 1)$ the limit is the zero function, which is continuous, but the sup criterion is not satisfied: since $x^{n} \to 1$ as $x \to 1^{-}$ , the supremum of $x^{n}$ on $[0, 1)$ equals $1$ for every $n$ , and it does not tend to $0$ . On $[0, a]$ the function $x^{n}$ is increasing, so the maximum is $a^{n}$ , which tends to $0$ because $0 < a < 1$ ; therefore the convergence is uniform on every interval $[0, a]$ with $a < 1$ .

Exercise 2

Study the pointwise and uniform convergence of $f_{n} (x) = arctan (n x)$ on $R$ and on $[δ, + \infty)$ with $δ > 0$ .

f (x) = ⎩ ⎨ ⎧ \frac{π}{2} 0 - \frac{π}{2} if x > 0 if x = 0 if x < 0

R : f discontinuous at 0 ⟹ convergence not uniform

x \geq δ : 0 \leq \frac{π}{2} - arctan (n x) \leq \frac{π}{2} - arctan (n δ)

M_{n} = x \geq δ sup f_{n} (x) - \frac{π}{2} = \frac{π}{2} - arctan (n δ) n \to + \infty 0 ⟹ uniform convergence on [δ, + \infty)

y = \arctan x

y = \arctan \left(4 x\right)

y = \arctan \left(20 x\right)

y = arctan (n x)

for

n = 1, 4, 20

: for large

n

the graph approaches

y = \frac{π}{2}

for

x > 0

and

y = - \frac{π}{2}

for

x < 0

(dashed lines), with the transition concentrated near the origin

For fixed $x > 0$ the argument $n x$ tends to $+ \infty$ , so $arctan (n x) \to \frac{π}{2}$ ; for $x < 0$ the limit is $- \frac{π}{2}$ , and at $x = 0$ every term equals $0$ . The pointwise limit is therefore discontinuous at the origin, while every $f_{n}$ is continuous, so the convergence is not uniform on $R$ . On the half-line $[δ, + \infty)$ the difference $\frac{π}{2} - arctan (n x)$ is positive and decreasing in $x$ , since the arctangent is increasing, so the sup is attained at the first point $x = δ$ . Since $n δ \to + \infty$ , the quantity $\frac{π}{2} - arctan (n δ)$ tends to $0$ , and the sup criterion guarantees uniform convergence on $[δ, + \infty)$ for every $δ > 0$ .

Exercise 3

Study the pointwise and uniform convergence of $f_{n} (x) = e^{- n x}$ on $(0, + \infty)$ and on $[δ, + \infty)$ with $δ > 0$ .

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

(0, + \infty) : M_{n} = x > 0 sup e^{- n x} = x \to 0^{+} lim e^{- n x} = 1 \neq \to 0 ⟹ convergence not uniform

[δ, + \infty) : M_{n} = x \geq δ sup e^{- n x} = e^{- n δ} n \to + \infty 0 ⟹ uniform convergence

For each fixed $x > 0$ the exponent $- n x$ tends to $- \infty$ , so $f_{n} (x) \to 0$ and the pointwise limit is the zero function. The function $e^{- n x}$ is decreasing in $x$ , therefore on $(0, + \infty)$ the supremum is found at the left endpoint: as $x \to 0^{+}$ the values tend to $1$ , so $M_{n} = 1$ for every $n$ and the convergence is not uniform. In this case the pointwise limit is continuous on $(0, + \infty)$ , so the example shows that the theorem on the continuity of the uniform limit does not have a converse. On $[δ, + \infty)$ monotonicity gives $M_{n} = e^{- n δ}$ , which tends to $0$ because $δ > 0$ is fixed, therefore the convergence is uniform on every half-line $[δ, + \infty)$ .