Exercise 1.14

Question

\def\since{\Leftarrow}
\def	hen{\Rightarrow}
\def\eg{	extit{e.g.} }
\def\ie{	extit{i.e.} }
\def\Def{	riangleq}
\def\N{\usualSet{N}}
\def\C{\usualSet{C}}
\def\minus{\insmall{-}}

Put $K =[0, 1]$ and define $\D_K$ as in Section 1.46. 
Show that the following three families of seminorms 
  (where $n = 0, 1, 2, \dots$) define the same topology on $\D_K$. 
If $D = d/dx$: 
%
  \begin{enumerate}
    \item{
      $\| D^n f \|_\infty = \sup\set{\left| D^n f(x)ight|}{\infty< x< \infty}$
    }
    \item{
      $\| D^n f \|_1 =\int_0^1 \left|D^n f(\varit{x}) ight| d\varit{x}$
    }
    \item{
      $\| D^n f \|_2 = \left\{
        \int_0^1 | D^n f(\varit{x}) |^2  d\varit{x} 
      ight\}^{1/2}.$
    }
  \end{enumerate}

Cordier · Accepted Answer

%
\begin{proof}%
Let us equipp $\D_K$ with the inner product %
%
  $\bra{f}\ket{g} = \int_0^1 f \, \bar{g}$, so that %
  $\bra{f}\ket{f} = \| f \|_2^2$. %
%
The following %
%
\begin{align}
  \int_0^1 1\,| D^n f| \leq \| 1 \|_2 \| D^n f \|_2 
\end{align}
%
is then a Cauchy-Schwarz inequality; see Theorem 12.2 of 	extit{Functional Analysis}. %
We so obtain %
%
\begin{align}\label{inequalities 1}
  \| D^n f \|_1 
    \leq 
  \| D^n f \|_2 
    \leq 
  \| D^n f \|_\infty 
    <\infty 
\end{align}
%
since $K$ has length $1$. %
%
Obviously, the support of $D^n f$ lies in $K$, hence the below equality %
%
\begin{align}\label{inequalities 2}
  |D^n f(x)|
    =
  \left|\int_{0}^x D^{n+1}fight|
    \leq 
  \int_{0}^x |D^{n+1}f|
    \leq 
  \|D^{n+1}f \|_1.
\end{align}
%
Take the supremum over all $|D^n f(x)|$: Combining (ef{inequalities 1}) %
with (ef{inequalities 2}) now reads as follows, %
%
\begin{align}\label{inequalities 3}
  \|D^n f\|_1 
    \leq 
  \|D^{n} f\|_2 
    \leq 
  \|D^{n} f\|_\infty 
    \leq 
  \|D^{n+1}f\|_1 
    \leq 
  \cdots < \infty.
\end{align}
%
Finally, put %
%
\begin{align}
  V^{(i)}_n         & 	riangleq \{f\in \D_K: \|f \|_i < 2^{\,\minus n}\}, \
  \mathscr{B}^{(i)} & 	riangleq \{V^{(i)}_n: n = 0, 1, 2, \dots\},
\end{align}
%
so that (ef{inequalities 3}) is mirrored by neighborhood inclusions, %
provided $i=1, 2, \infty$:
%
\begin{align}\label{inclusions}
  V^{(1)}_n
    \supseteq  
  V^{(2)}_n 
    \supseteq  
  V^{(\infty)}_n 
    \supseteq  
  V^{(1)}_{n+1} 
    \supseteq  
  \cdots .
\end{align}
%
Their subchains $V^{(i)}_n\supseteq V^{(i)}_{n+1}$ turn $\mathscr{B}^{(i)}$ %
into a local base of a topology $	au_i$. 
The whole chain (ef{inclusions}) then forces %
%
\begin{align}
  	au_1 \subseteq 	au_2 \subseteq 	au_\infty \subseteq 	au_1;
\end{align}
%
which achieves the proof.
\end{proof}
% END

Exercise 1.14

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Exercise 1.14

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