Let S be the set of all measurable functions on [0,1]. Then the set O
of all functions in S equal to 0 a.e on [0,1] makes an additive subgroup
of a commutative group S. Show that S/O with the distance
p(f,g)=indefinite integral(0,1)((|f(x)-g(x)|)/(1+|f(x)-g(x)))dx
is a metric space. Show that a sequence {fn}n>=1 converges to zero in
(S/O,p) iff {fn}n>=1 converges to zero measure.
7 answers
Sorry, can not help, hope someone can.
I can't help either
this question has me "stumped" too, have not seen one like that before.
Wow. Algebra and Real Analysis. I hope this isn't a high-school problem!
But anyway, to get you started just look at the definition of metric spaces.
In this case, note that:
(a) p(f, g) >= 0 (this should be clear), and as |f - g| = 0 iff f == g.
(b) p(f, g) = p(g, f) (because of the way modulus works).
Finally, see whether:
(c) Triangle inequality holds (check this!)
(Triangle inequality: |a + b| <= |a| + |b|, and in this case use the L1 norm (your p) instead of simple modulus, and the variables are functions).
If that stuff holds in S mod O, then you're golden.
In this case, note that:
(a) p(f, g) >= 0 (this should be clear), and as |f - g| = 0 iff f == g.
(b) p(f, g) = p(g, f) (because of the way modulus works).
Finally, see whether:
(c) Triangle inequality holds (check this!)
(Triangle inequality: |a + b| <= |a| + |b|, and in this case use the L1 norm (your p) instead of simple modulus, and the variables are functions).
If that stuff holds in S mod O, then you're golden.
Whoops, I meant
p(f, h) <= p(f, g) + p(g, h)
for (c).
p(f, h) <= p(f, g) + p(g, h)
for (c).
thank you but what will ý do at the end?
1 answer