Question #170286

Show that a function of bounded variation on [a,b] is bounded therein


1
Expert's answer
2021-03-15T08:10:58-0400

 Let f be a function of bounded variation on the interval [a,b].

Then there exists a positive real number M>0 such that for all partitions P∈ P[a,b]

 we have that:


Vf(P)=Σk=1nf(xk)f(xk1)MV_f(P)=\Sigma_{k=1}^n|f(x_k)-f(x_{k-1})|\leq M

For all x∈[a,b] consider the partition P={a,x,b} (where P={a,b} if x=a or x=b).

Then:


Vf({a,x,b})=f(b)f(x)+f(x)f(a)MV_f(\{a,x,b\})=|f(b)-f(x)|+|f(x)-f(a)|\leq M

Hence we have that f(x)f(a)M,|f(x)-f(a)|\leq M, so for all x∈[a,b] we have that f(x)f(a)+M|f(x)|\leq |f(a)|+M ,

So, f is bounded on [a,b]





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