Map from Set of Continuous Functions on Interval to Set of their Integrals

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Theorem

Let $\Bbb I = \closedint a b$ be a closed interval on the set of real numbers $\R$ such that $a < b$.

Let $A$ denote the set of all continuous real functions $f: \Bbb I \to \R$.

Let $B \subseteq A$ denote the set of all functions differentiable on $\Bbb I$ whose derivative is continuous on $\Bbb I$.

Let $C \subseteq B$ denote the subset of $B$ which consists of all elements of $B$ such that $\map f a = 0$.


For each $f \in A$, let $h$ denote the mapping defined as:

$\map {\paren {\map h f} } x = \displaystyle \int_a^x \map f t \rd t$


Then:

$h: A \to C$


Proof

Let $f \in A$ be an arbitrary continuous real function $f: \Bbb I \to \R$.

From Continuous Function is Riemann Integrable, $\map {\paren {\map h f} } x$ exists and is continuous on $\Bbb I$.


Let $x = a$.

Then we have:

\(\displaystyle \map {\paren {\map h f} } x\) \(=\) \(\displaystyle \int_a^x \map f t \rd t\) Definition of $h$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \map {\paren {\map h f} } a\) \(=\) \(\displaystyle \int_a^a \map f t \rd t\)
\(\displaystyle \) \(=\) \(\displaystyle 0\) Definite Integral on Zero Interval

Thus $h$ is a continuous real function on $\Bbb I$ such that $\map h a = 0$.


Further, by definition, $\int \map f t \rd t$ is the primitive of $f$, and exists by dint of the above.

It follows from the Fundamental Theorem of Calculus that $f = \map {\dfrac \d {\d x} } h$ and so $\map h x$ is functions differentiable on $\Bbb I$.

Thus it is shown that $C \subseteq B \subseteq A$ and the result follows.

$\blacksquare$


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