I think I understand now, so you are saying that if N=x and n>3.5 then f_n(3.5)=0?
Right (but you want N to be natural, so in this case N=4 or N=3 works)! So if instead I asked you to do it for 10.3, what would you say to me? How bout 11? In general, if I gave you an x and asked you to find the N, what would you say?
Right (but you want N to be natural, so in this case N=4 or N=3 works)! So if instead I asked you to do it for 10.3, what would you say to me? How bout 11? In general, if I gave you an x and asked you to find the N, what would you say?
So to structure the proof, I'd write this Fix x in real. Fix epsilon >0. Take any natural N bigger than or equal to x. Then for n>N we have |f_n(x)-0|=|0-0|<epsilon. Thus f_n(x) converges pointwise to the 0 function (notice the convergence is not uniform. See if you can find a sequence of functions for which f_n converges uniformly to the 0 function but the integrals do not converge to 0).
So to structure the proof, I'd write this Fix x in real. Fix epsilon >0. Take any natural N bigger than or equal to x. Then for n>N we have |f_n(x)-0|=|0-0|<epsilon. Thus f_n(x) converges pointwise to the 0 function (notice the convergence is not uniform. See if you can find a sequence of functions for which f_n converges uniformly to the 0 function but the integrals do not converge to 0).
Can I just check, to show that f_n is bounded is is just the fact that it can't be more than 1?
Ah ok so would it be enough to say that as n tends to infinity, the interval becomes so small that the indicator will be always zero and so the function converges to zero?
Sort of. It's more that if you pick any point and let n be large enough, then the function evaluated at that point will eventually be zero.
Not sure how you can say f_n can't be more than 1? Unless there's some assumption you're missing out, that's clearly not true.
You are right, I forgot that I was looking at a sequence of functions rather than a single function. So if I am trying to show that f_n is a bounded sequence of functions, am I trying to find an M such that for any n>=1 fn will always be less than it?
You are right, I forgot that I was looking at a sequence of functions rather than a single function. So if I am trying to show that f_n is a bounded sequence of functions, am I trying to find an M such that for any n>=1 fn will always be less than it?
It's somewhat dangerous to speak of 'bounded' without defining the metric you're using.
For the function under discussion:
n→∞lim0≤x≤1sup∣fn(x)∣=∞, so f isn't bounded under the sup norm.
At the same time,
limn→∞∫∣fn(x)∣dx=1 so f is bounded under the L1 norm.
More generally, one of the key points with these questions is to clearly distinguish (in your mind, and ideally in your answer!) the difference between what's happening at a particular point, and what's happening over the range of the entire function. It's something you seem to be confusing a lot in your posts here.
Can I check, I know that the bounded convergence theorem doesn't apply to the second example as the real numbers are not a bounded interval. However, for the first example does the theorem not apply because the function is not bounded?