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When a real-valued function of one variable is approximated by its $n$th degree Taylor polynomial, the remainder is estimated using the Alexiewicz and Lebesgue $p$-norms in cases where $f^{\left(n\right)}$ or $f^{(n+1)}$ are Henstock-Kurzweil integrable. When the only assumption is that $f^{\left(n\right)}$ is Henstock-Kurzweil integrable then a modified form of the $n$th degree Taylor polynomial is used. When the only assumption is that $f^{\left(n\right)}\in C^0$ then the remainder is estimated by applying the Alexiewicz norm to Schwartz distributions of order 1.

Let ${ℬ}_c$ denote the real-valued functions continuous on the extended real line and vanishing at $-\infty$. Let ${ℬ}_r$ denote the functions that are left continuous, have a right limit at each point and vanish at $-\infty$. Define ${𝒜}_c^n$ to be the space of tempered distributions that are the $n$th distributional derivative of a unique function in ${ℬ}_c$. Similarly with ${𝒜}_r^n$ from ${ℬ}_r$. A type of integral is defined on distributions in ${𝒜}_c^n$ and ${𝒜}_r^n$. The multipliers are iterated integrals of functions of bounded variation. For each $n\in \mathbb{N}$, the spaces...

If $f$ is a Henstock-Kurzweil integrable function on the real line, the Alexiewicz norm of $f$ is $\parallel f\parallel ={sup}_I\left|{\int }_{I}f\right|$ where the supremum is taken over all intervals $I\subset ℝ$. Define the translation ${\tau }_x$ by ${\tau }_{x}f\left(y\right)=f(y-x)$. Then $\parallel {\tau }_{x}f-f\parallel$ tends to $0$ as $x$ tends to $0$, i.e., $f$ is continuous in the Alexiewicz norm. For particular functions, $\parallel {\tau }_{x}f-f\parallel$ can tend to 0 arbitrarily slowly. In general, $\parallel {\tau }_{x}f-f\parallel \ge \text{osc}f\left|x\right|$ as $x\to 0$, where $\text{osc}f$ is the oscillation of $f$. It is shown that if $F$ is a primitive of $f$ then $\parallel {\tau }_{x}F-F\parallel \le \parallel f\parallel \left|x\right|$. An example shows that the function $y\mapsto {\tau }_{x}F\left(y\right)-F\left(y\right)$ need not be in $L^1$. However, if $f\in L^1$ then $\parallel {\tau }_x{F-F\parallel }_1\le {\parallel f\parallel }_1\left|x\right|$....