$Epsilon delta 2$

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$Epsilon delta 2$

7 Let f : R 7→ R be a function with

sup x∈R |f(x)| ≤ B

for a finite number B. Show by an ε − δ argument that the function F : R 7→ R given by

F(x) = integral from 0-x f(s)ds

is continuous at each point x0 ∈ R.

Calculus

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Let
\[F(x)=\int_0^{x}f(s)ds.\]
Then
\[|F(x)-F(y)|=|\int_0^{x}f(s)ds -\int_0^{y}f(s)ds |=|\int_x^{y}f(s)ds|\]
\[\leq |\int_x^{y}|f(s)|ds| \leq B |\int_x^{y}ds|=B|x-y|.\]
Thus
\[|F(x)-F(y)| \leq B |x-y|.\]
Now let $\epsilon >0$ and $\delta= \frac{\epsilon}{B}$. Then for $|x-y|<\delta$ we have
\[|F(x)-F(y)| \leq B |x-y|<B \delta=B \frac{\epsilon}{B}=\epsilon.\]
Hence $F$ is continuous for all $x \in \R$. The above argument indeed shows that $F$ is also uniformly continuous on $\R$.

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$Epsilon delta 2$

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