- #1
GridironCPJ
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Q: Suppose that the oscillation ω_f (x) of a function f is smaller than η at each point x of an interval [c,d]. Show that there must be a partition π of [c,d] s.t. the oscillation
ωf([x_(k-1),x_k ])<η
on each member of the partition.
My solution (Rough sketch):
This condition on x is local, so it must be true for a δ-neightborhood of x s.t. ωf(δ(x))<η. Now take a partition s.t. each subinterval [x_(k-1),x_k ]<δ. Thus, each subinterval is less than the δ from the δ-neightborhood of x, so then
ωf([x_(k-1),x_k ])[itex]\leq[/itex]ωf(δ(x))<η. QED
Is this logic too sloppy? If so, does anyone have any suggestions as to a more proper way to prove this?
ωf([x_(k-1),x_k ])<η
on each member of the partition.
My solution (Rough sketch):
This condition on x is local, so it must be true for a δ-neightborhood of x s.t. ωf(δ(x))<η. Now take a partition s.t. each subinterval [x_(k-1),x_k ]<δ. Thus, each subinterval is less than the δ from the δ-neightborhood of x, so then
ωf([x_(k-1),x_k ])[itex]\leq[/itex]ωf(δ(x))<η. QED
Is this logic too sloppy? If so, does anyone have any suggestions as to a more proper way to prove this?