Question #288694

Show that f is continuous on a,b and if f' exists is bounded on interior, says f'x less than A for all x in a,b then the f is of bounded variation

1
Expert's answer
2022-01-19T17:53:23-0500

Theorem: If f is continuous on [a, b] and if ff^{\prime} exists and is bounded in the interior, say f(x)A x(a,b)\left|f^{\prime}(x)\right| \leq A \forall\ x \in (a, b) , then f is of bounded variation on [a, b].


Proof: Applying the Mean-Value Theorem, we have

Δfk=f(xk)f(xk1)=f(tk)(xkxk1), where tk(xk1,xk).\Delta f_{k}=f\left(x_{k}\right)-f\left(x_{k-1}\right)=f^{\prime}\left(t_{k}\right)\left(x_{k}-x_{k-1}\right), \quad \text { where } t_{k} \in\left(x_{k-1}, x_{k}\right) .

This implies

k=1nΔfk=k=1nf(tk)ΔxkAk=1nΔxk=A(ba).\sum_{k=1}^{n}\left|\Delta f_{k}\right|=\sum_{k=1}^{n}\left|f^{\prime}\left(t_{k}\right)\right| \Delta x_{k} \leq A \sum_{k=1}^{n} \Delta x_{k}=A(b-a) .


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