Boundedness
Key Questions
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Answer:
Boundedness is about having finite limits. In the context of values of functions, we say that a function has an upper bound if the value does not exceed a certain upper limit. More...
Explanation:
Other terms used are "bounded above" or "bounded below".
For example, the function
#f(x) = 1/(1+x^2)# is bounded above by#1# and below by#0# in that:#0 < f(x) <= 1# for all#x in RR# graph{1/(1+x^2) [-5, 5, -2.5, 2.5]}
The function
#exp:x -> e^x# is bounded below by#0# (or you can say has#0# as a lower bound), but is not bounded above.#0 < e^x < oo# for all#x in RR# graph{e^x [-5.194, 4.806, -0.74, 4.26]}
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Answer:
A continuous function defined on a closed interval has an upper (and lower) bound.
Explanation:
Probably the simplest boundedness theorem states that a continuous function defined on a closed interval has an upper (and lower) bound.
Proof by contradiction
Suppose
#f(x)# is defined and continuous on a closed interval#[a, b]# , but has no upper bound.Then:
#AA n in NN, EE x_n in [a, b] : f(x_n) > n# Since the sequence of
#x_n# 's lies in a bounded interval, it is dense at some point in the closure of the interval. Since the interval is closed, that must be at some point#c# actually in the interval#[a, b]# .Since the sequence of
#x_n# 's is dense at#c# , there is some monotonically increasing sequence#n_k in NN# such that#x_(n_k) -> c# as#k->oo# .Now
#f(x)# is continuous at#c# , so:#lim_(x->c) f(x) = f(c)# which is bounded.
But:
#lim_(k->oo) x_(n_k) = c" "# and#" "lim_(k->oo) f(x_(n_k)) = oo# is unbounded.
...contradiction.
So there is no such
#f(x)# lacking upper (or lower) bound. -
If the function is unbounded, the graph would progress to infinity, in some direction(s).
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Not all functions are bounded.
The simplest counter example would be
the identity function#f(x) = x#
which is defined for all values of#x# and can generate any value for#f(x)# A slightly less trivial counter example would be
the cubing function#f(x) = x^3#