How about f(x) = floor(x)? (On, say, [0,1].) It's monotone and therefore of bounded variation, but is not Lipschitz continuous (or even continuous).
A circle
a circle
NO
The answer is yes is and only if da limit of the sequence is a bounded function.The suficiency derives directly from the definition of the uniform convergence. The necesity follows from making n tend to infinity in |fn(x)|
If the range is the real numbers, it has a lower bound (zero) but no upper bound.
No. y = 1/x is continuous but unbounded.
That depends! The identity operator must map something from a space X to a space Y. This mapping might be continuous - which is the case if the identify operator is bounded - or discontinuous - if the identity operator is unbounded.
A signal is bounded if there is a finite value such that the signal magnitude never exceeds , that is for discrete-time signals, or for continuous-time signal (Source:Wikipedia)
A circle
a circle
NO
GREEN'S THEOREM: if m=m(x,y) and n= n(x,y) are the continuous functions and also partial differential in a region 'r' of x,y plane bounded by a simple closed curve c. DIVERGENCE THEOREM: if f is a vector point function having continuous first order partial derivatives in the region v bounded by a closed curve s
What is the area bounded by the graph of the function f(x)=1-e^-x over the interval [-1, 2]?
The answer is yes is and only if da limit of the sequence is a bounded function.The suficiency derives directly from the definition of the uniform convergence. The necesity follows from making n tend to infinity in |fn(x)|
A function whose upper bound would have attained its upper limit at a bound. For example, f(x) = x - a whose domain is a < x < b The upper bound is upper bound is b - a but, because x < b, the bound is never actually attained.
the abdominal what? ... the abdomen is that part of the human body inferiorly continuous with the pelvic area, and bounded superiorly by the diaphragm.
A function may have a finite number of discontinuities and still be integrable according to Riemann (i.e., the Riemann integral exists); it may even have a countable infinite number of discontinuities and still be integrable according to Lebesgue. Any function with a finite amount of discontinuities (that satisfies other requirements, such as being bounded) can serve as an example; an example of a specific function would be the function defined as: f(x) = 1, for x < 10 f(x) = 2, otherwise