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Accumulation Points

Consider the subset A of C([0,1]) consisting of continuous function f with f(0)=f(1)=0

In [latex ](C([0,1]),||\cdot||_1 ) determine whether the following are accumulation points of the set A

1)

g_1(t)=0

g_2(t)=t

How would i go about solving this type of question?

Reply 1

Mark13

Original post by gemma331

Consider the subset A of C([0,1]) consisting of continuous function f with f(0)=f(1)=0

In [latex ](C([0,1]),||\cdot||_1 )

determine whether the following are accumulation points of the set A

1)

g_1(t)=0

g_2(t)=t

How would i go about solving this type of question?

Do you know the definition of an accumulation point? Which of g_1 and g_2 intuitively seems like it could be an accumulation point?

Reply 2

gemma331

Original post by Mark13

Do you know the definition of an accumulation point? Which of g_1 and g_2 intuitively seems like it could be an accumulation point?

The one i think is more likely to be an accumulation point is g_1 but im really not sure how i go about solving this... the examples i have previously done.. in previous work i have been given an

a_n

and find the accumulation points through

b_n

where

b_n = |a_n|^\frac{1}{n}

Im not really sure where i go with this question, but proving which one is and proving why the other isnt.

Reply 3

Mark13

Original post by gemma331

The one i think is more likely to be an accumulation point is g_1 but im really not sure how i go about solving this... the examples i have previously done.. in previous work i have been given an

a_n

and find the accumulation points through

b_n

where

b_n = |a_n|^\frac{1}{n}

Im not really sure where i go with this question, but proving which one is and proving why the other isnt.

The first thing you need to do is go back to your lecture notes/text book and get familiar with what an accumulation point is - feel free to post any questions you have about the definition up here.

Reply 4

gemma331

Original post by Mark13

The first thing you need to do is go back to your lecture notes/text book and get familiar with what an accumulation point is - feel free to post any questions you have about the definition up here.

I know I have to find two sequences of functions in A one t hat converges to

g_1(t)

and the other to

g_2(t)

but I really don't know how I go about doing ththis please help!!

Reply 5

Mark13

Original post by gemma331

I know I have to find two sequences of functions in A one t hat converges to

g_1(t)

and the other to

g_2(t)

but I really don't know how I go about doing ththis please help!!

Think about an arbitrary function f in A. Then we know that f is continuous, and f(0)=f(1)=0. Also,

||f - g_2||_{\infty} = \sup_{0 \leq x \leq 1} |f(x) - g_2(x)| \geq |f(1)-g_2(1)|

.

Can you see how to use this to show that g_2 isn't an accumulation point of A?

g_1 is an accumulation point of A - to show this you need to find a sequence of continuous functions, each of which takes the value 0 at x=0 and x=1, which converge to g_1 in the uniform norm.

Reply 6

gemma331

Original post by Mark13

Think about an arbitrary function f in A. Then we know that f is continuous, and f(0)=f(1)=0. Also,

||f - g_2||_{\infty} = \sup_{0 \leq x \leq 1} |f(x) - g_2(x)| \geq |f(1)-g_2(1)|

.

Can you see how to use this to show that g_2 isn't an accumulation point of A?

I still don't understand how this shows g_2 isn't an accumulation point :s

Reply 7

Mark13

Original post by gemma331

I still don't understand how this shows g_2 isn't an accumulation point :s

If g_2 is an accumulation point, then there is a sequence of functions f_n in A such that

||f_n - g_2||_{\infty} \rightarrow 0

. If you can show that for any function f in A, we have

||f - g_2|| \geq 1

, then no such sequence f_n can exist, so g_2 is not an accumulation point.

Reply 8

ghostwalker

Original post by Mark13

...

Just noting that you seem to have switched from the 1-norm to the infinity-norm.

Reply 9

Mark13

Original post by ghostwalker

Just noting that you seem to have switched from the 1-norm to the infinity-norm.

Thanks for pointing that out, I need to read things more carefully :redface:

In that case, both functions are accumulation points, and to show that a given function g is an accumulation point, you need to find a sequence of functions f_n such that

||f_n - g||_1 = \int_0^1|f_n(x)-g(x)|dx \rightarrow 0

.

Often a good way to find a sequence of functions converging to the required limit is to think graphically. For example, the sequence of functions f_n defined by:

f_n agrees with g_2 for all x satisfying

0 \leq x \leq 1-1/n

, and then linearly intepolating between the points (1-1/n,1-1/n) and (1,0) after that (it's a lot easier to draw this function than give an expression for it)

will converge to g_2 in the 1-norm.

(edited 11 years ago)