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The set of all bounded sequences forms the sequence space. [ citation needed ] The definition of boundedness can be generalized to functions f : X → Y {\displaystyle f:X\rightarrow Y} taking values in a more general space Y {\displaystyle Y} by requiring that the image f ( X ) {\displaystyle f(X)} is a bounded set in Y {\displaystyle Y} .
In its simplest form, it says that a non-decreasing bounded-above sequence of real numbers ... converges to its smallest upper bound, its supremum. Likewise, a non-increasing bounded-below sequence converges to its largest lower bound, its infimum. In particular, infinite sums of non-negative numbers converge to the supremum of the partial sums ...
Proof: (closed and bounded implies sequential compactness) Since is bounded, any sequence {} is also bounded. From the Bolzano-Weierstrass theorem, {} contains a subsequence converging to some point .
Corollary — If a sequence of bounded operators () converges pointwise, that is, the limit of (()) exists for all , then these pointwise limits define a bounded linear operator . The above corollary does not claim that T n {\displaystyle T_{n}} converges to T {\displaystyle T} in operator norm, that is, uniformly on bounded sets.
Every uniformly convergent sequence of bounded functions is uniformly bounded. The family of functions f n ( x ) = sin n x {\displaystyle f_{n}(x)=\sin nx} defined for real x {\displaystyle x} with n {\displaystyle n} traveling through the integers , is uniformly bounded by 1.
Since every convergent sequence is bounded, is a linear subspace of . Moreover, this sequence space is a closed subspace of ℓ ∞ {\displaystyle \ell ^{\infty }} with respect to the supremum norm , and so it is a Banach space with respect to this norm.
Bounded poset, a partially ordered set that has both a greatest and a least element; Bounded set, a set that is finite in some sense Bounded function, a function or sequence whose possible values form a bounded set; Bounded set (topological vector space), a set in which every neighborhood of the zero vector can be inflated to include the set
Since the sequence is uniformly bounded, there is a real number M such that |f n (x)| ≤ M for all x ∈ S and for all n. Define g(x) = M for all x ∈ S. Then the sequence is dominated by g. Furthermore, g is integrable since it is a constant function on a set of finite measure. Therefore, the result follows from the dominated convergence ...