Prove that function sequence converges uniformly to 0

[velvet lion]

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[dense ember]

what is the sup of f_n(x) on [a, b] ?

[vagrant terrace]

If by sup you mean the supremum, then b, because (f_n) is a sequence of monotonically increasing functions.

[dense ember]

f_n(b)

now it is not hard to show that $\sup_{x \in [a, b]} f_n(x) \to 0$

[ocean tartan]

Rion

[vagrant terrace]

Since f(a)>= 0 and f(a)<=f(x) by monotonicity for all x∈[a,b], f(x)>= 0 finally applies.
By lim(n)→∞ f(b) = 0 and f(a)<=f(x)<=f(b), it finally follows that lim (n)→∞ f(x)=0.
Or have I misinterpreted something?

[dense ember]

uniform convergence to 0 is equivalent to the convergence of the sup to 0

[dense ember]

if you think hard about it, the definition is equivalent to the convergence of the sup

because the sup upper-bounds the individual difference |f_n(x) - f(x)|

you can try using the raw definition but it is much more tedious

[vagrant terrace]

Because if sup<epsilon -> all x < epsilon ?

[dense ember]

Well informally yes

if $\sup_{t \in D} \lvert f_n(t) - f(t) \rvert < \epsilon$, then for all $x \in D$, \lvert f_n(x) - f(x) \rvert < \epsilon$

[ocean tartan]

Rion
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[dense ember]

for the simple and good reason that $\lvert f_n(x) - f(x) \rvert \leq \sup_{t \in D} \lvert f_n(t) - f(t) \rvert$

[ocean tartan]

Rion

[dense ember]

Conversely, if this is verified

if for all $x \in D, \lvert f_n(x) - f(x) \rvert < \epsilon$

[ocean tartan]

Rion

[dense ember]

you'd certainly have $\sup_{t \in D} \lvert f_n(t) - f(t) \rvert \leq \epsilon$

[ocean tartan]

Rion

[dense ember]

that's not good enough

all you have proven here is that you have pointwise convergence to zero

not necessarily uniform

if you want, we can go back to your initial definition (though it is a bit inefficient)

you know that: f_n(b) tends to 0

so let epsilon > 0, and let n_0 such that for all n>= n0, f_n(b) <= epsilon

does this n_0 make it work?

To show that this is equivalent to the sup converging to zero

let epsilon > 0

you apply the property with epsilon/2 : let n_0 such that for all n >= n_0,

for all x in D, | f_n(x) - f(x) | < epsilon/2

in particular, taking the sup (by closure), you get:
sup_t |f_n(t) - f(t)| <= epsilon/2 < epsilon

so: for all epsilon > 0, there exists n_0, such that for all n >= n_0, sup_t |f_n(t) - f(t)| < epsilon

and the other direction is already proven above

so from now on, to show uniform convergence, you can simply look at the sup

[vagrant terrace]

t is variable for a number in [a,b], or not?

[dense ember]

yes

well it's the local variable used by sup

Well let me know if you need clarification on anything

[vagrant terrace]

Is there a particular reason why we have chosen exactly ε/2? What exactly speaks against using only ε and would it also have been possible, for example, to use (3/4)ε?

[dense ember]

Just needed it to be strictly less than epsilon

(3/4) epsilon works too

[vagrant terrace]

Is it also | f_n(x) - f(x) | <= ε/2 here or only < ε/2

[dense ember]

According to the property, it's <

When you move to the sup, < becomes <=

That's why I needed something less than epsilon

[vagrant terrace]

For example, does supx∈[a,b]∣fn(x)∣ mean “Find the largest value of |f_n(x)| when x runs over the entire interval [a,b]”? Or did I misunderstand the notation?

[dense ember]

$\sup_{t \in D} g(t) = \sup { g(t) : t \in D}$

[ocean tartan]

Rion

[dense ember]

The least upper-bound

Supremum

[vagrant terrace]

Ok, thank you. I think I've understood everything now.

[vagrant terrace]

+close

[knotty acorn]

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[knotty acorn]

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