anonymous
  • anonymous
Determine whether the following are subspaces of C[-1,1]: (c) The set of continous nondecreasing functions on [-1,1] I already know how to prove that something is a subspace, but the problem here is in the term nondecreasing which i don't know how to turn it into a "such that" statement to test it
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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Zarkon
  • Zarkon
\[f(x)=x\] \[(-1)\cdot f(x)=-x\]
KingGeorge
  • KingGeorge
Try: A function \(f(x)\) is non-decreasing on an interval \(I\) if \(f(a)\leq f(b)\) for all \(a, b \in I\) such that \(a < b\).
anonymous
  • anonymous
i'm not sure if you can see what I wrote, i'm not good with the equation typing buton

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KingGeorge
  • KingGeorge
What was cut off at the end? I can only read "\(\alpha f(x)\) might be decre...." By the way, if you want put \(\LaTeX\) in line, type "\.( the stuff you want the be in math format \.)" Just remove the periods.
anonymous
  • anonymous
"\ ( if i want to test for closure under scalar multiplication, i will do the following if f in C[-1,1] then alpha f in C[-1,1] also f(x) is non-decreasing means that alpha f(x) might be decreasing when alpha is negative or so ? \)"
anonymous
  • anonymous
this would mean that f is not in c[-1,1] ?
KingGeorge
  • KingGeorge
I believe so. Since it isn't closed under scalar multiplication, not a subspace.
Zarkon
  • Zarkon
hence my counter example above.
anonymous
  • anonymous
from your example, we can write the statement "is non-decreasing" as "f(x) = x is non decreasing" ?
KingGeorge
  • KingGeorge
Yes. Assume \(f(x)\) is non-decreasing, then \((-1)f(x)\) is decreasing.
anonymous
  • anonymous
ok, I see. thanks

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