Question #170332

If f is continuous on [a,b] and f' is bounded in(a,b) prove that f is of bounded variation on [a,b]


Expert's answer

Since f′f' is bounded,

∃ M>0∋∣f′(α)∣≤M ∀α∈[a,b]\exist \, M>0 \ni | f'(\alpha)| \leq M \, \forall \alpha \in [a,b]

Let P={x0,x1,...,xn}P = \{ x_0,x_1,...,x_n \} be a partition of [a,b][a,b]

Then by the Mean Value Theorem, we choose α∈(xi−1,xi) ∋\alpha \in (x_{i-1},x_i) \, \ni f(xi)−f(xi−1)=f′(α)(xi−xi−1)f(x_i) - f(x_{i-1}) = f'(\alpha)(x_i - x_{i-1})

Therefore∣f(xi)−f(xi−1)∣=∣f′(α)∣∣xi−xi−1∣≤M∣xi−xi−1∣|f(x_i) - f(x_{i-1})| = |f'(\alpha)||x_i - x_{i-1}| \leq M|x_i - x_{i-1}|

Hence, we have

∑i=1n∣f(xi)−f(xi−1)∣≤M∑i=1n∣xi−xi−1∣=M(b−a)\sum_{i=1}^{n}|f(x_i) - f(x_{i-1})| \leq M \sum_{i=1}^{n} |x_i - x_{i-1}| = M(b-a)

V(f,a,b,P)≤M(b−a)V(f,a,b,P) \leq M(b-a)

Since M(b−a)>0M(b-a) > 0 we have that f is of bounded variation.


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