Flashcards example
作者:
Uploaded from ShareLaTeX
最近上传:
6 年前
许可:
Other (as stated in the work)
摘要:
Example using the flashcards class.
This example was originally published on ShareLaTeX and subsequently moved to Overleaf in October 2019.
Example using the flashcards class.
This example was originally published on ShareLaTeX and subsequently moved to Overleaf in October 2019.
\documentclass[avery5388,grid,frame]{flashcards}
\cardfrontstyle[\large\slshape]{headings}
\cardbackstyle{empty}
\begin{document}
\cardfrontfoot{Functional Analysis}
\begin{flashcard}[Definition]{Norm on a Linear Space \\ Normed Space}
A real-valued function $||x||$ defined on a linear space $X$, where
$x \in X$, is said to be a \emph{norm on} $X$ if
\smallskip
\begin{description}
\item [Positivity] $||x|| \geq 0$,
\item [Triangle Inequality] $||x+y|| \leq ||x|| + ||y||$,
\item [Homogeneity] $||\alpha x|| = |\alpha| \: ||x||$,
$\alpha$ an arbitrary scalar,
\item [Positive Definiteness] $||x|| = 0$ if and only if $x=0$,
\end{description}
\smallskip
where $x$ and $y$ are arbitrary points in $X$.
\medskip
A linear/vector space with a norm is called a \emph{normed space}.
\end{flashcard}
\begin{flashcard}[Definition]{Inner Product}
Let $X$ be a complex linear space. An \emph{inner product} on $X$ is
a mapping that associates to each pair of vectors $x$, $y$ a scalar,
denoted $(x,y)$, that satisfies the following properties:
\medskip
\begin{description}
\item [Additivity] $(x+y,z) = (x,z) + (y,z)$,
\item [Homogeneity] $(\alpha \: x, y) = \alpha (x,y)$,
\item [Symmetry] $(x,y) = \overline{(y,x)}$,
\item [Positive Definiteness] $(x,x) > 0$, when $x\neq0$.
\end{description}
\end{flashcard}
\begin{flashcard}[Definition]{Linear Transformation/Operator}
A transformation $L$ of (operator on) a linear space $X$ into a linear
space $Y$, where $X$ and $Y$ have the same scalar field, is said to be
a \emph{linear transformation (operator)} if
\medskip
\begin{enumerate}
\item $L(\alpha x) = \alpha L(x), \forall x\in X$ and $\forall$
scalars $\alpha$, and
\item $L(x_1 + x_2) = L(x_1) + L(x_2)$ for all $x_1,x_2 \in X$.
\end{enumerate}
\end{flashcard}
\end{document}