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Let $P[0,1]$ be the space of real polynomial function on$[0,1]$ with $\|p\|= sup\{|p(x)|:0\leq x \leq 1\}. T(p)(x)=\frac{d}{dx}(p(x)).$ Then it is clear that $T$ is not a bounded operator. How to discuss the closeness of this operator. Is it close or not? If we consider the same operator on the space $C^{1}[0,1]$ then it is closed. But I am confused on the space $P[0,1]$. Please help me. Thanks.

1 Answers1

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If you consider ${d\over {dx}}:P[0,1]\rightarrow P[0,1]$ it is closed.

Let $p_n=\sum_ia_n^ix^i$ be a family of polynomial which converges towards the polynomial $p=\sum_ib^ix^i$. The sequence $(a_n^i)$ converges towards $b^i$. This implies that the sequence $(ia_n^i)$ converges towards $ib^i$ and $\lim_n {d\over {dx}}p_n={d\over{dx}}p$.

If you consider ${d\over{dx}}:C[0,1]\rightarrow C[0,1]$ whose domain is $P[0,1]$ the result is not true. Consider $e^x=\sum_{i\geq 0}{x^i\over{i!}}$. Write $p_n(x)=\sum_{i=0}^{i=n}{x^i\over{i!}}$; $e^x=lim_np_n$ and $lim_n{d\over{dx}}p_n=e^x$. You don't have $e^x\in P[0,1]$.

Tsemo Aristide
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  • Same as that for $C^{1}[0,1]$ ?? –  Aug 15 '16 at 18:54
  • This is not the definition of a closed linear operator. What are the Banach spaces? – zhw. Aug 15 '16 at 18:55
  • If you consider $P[0,1])$ has a subset of $C[0,1]$ then ${d\over {dx}}$ may not be closed, since a sequence of polynomials does not converge towards a polynomial necessarily. – Tsemo Aristide Aug 15 '16 at 18:57
  • Branch spaces are $C^{1} [0,1]$ ??? –  Aug 15 '16 at 18:57
  • See this definition: https://en.wikipedia.org/wiki/Unbounded_operator#Closed_linear_operators – zhw. Aug 15 '16 at 18:59
  • Explain why you believe that the answer here is not true – Tsemo Aristide Aug 15 '16 at 18:59
  • What is shown here respect the definition that you quoted, if $p_n$ converges towads $p$ and ${d\over {dx}}p_n$ converges towards $q$ since the domain of the operator is $P[0,1]$ and its image is also $P[0,1]$, $p,q$ are polynomials. – Tsemo Aristide Aug 15 '16 at 19:02
  • No it's not. Where are the Banach spaces $X,Y$? – zhw. Aug 15 '16 at 19:05
  • Can we consider the upper Banch space as $C^{1} [0,1]$ ??? –  Aug 15 '16 at 19:05
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    One more thing sequence of polynomial may not converges to polynomial... –  Aug 15 '16 at 19:07
  • @zhw : If $X$, $Y$ are normed spaces and $T : \mathcal{D}(T) \subseteq X \rightarrow Y$ is linear with a linear domain, then $T$ is closed iff the graph of $T$ is a closed subspace of $X\times Y$. Equivalently, if ${ x_n } \subseteq\mathcal{D}(T)$ converges in $X$ to $x$ and ${ Tx_n }$ converges in $Y$ to $y$, then $x\in\mathcal{D}(T)$ and $Tx=y$. Tsemo has pointed out an interesting case where $T$ is defined on polynomials. – Disintegrating By Parts Aug 15 '16 at 21:06
  • @TrialAndError how a sequence of polynomial must converge to a polynomial ? –  Aug 16 '16 at 03:30
  • @Dahiya : A sequence ${ \alpha_{0,n} + \alpha_{1,n} x + \cdots + \alpha_{N,n} x^N }$ converges if the coefficient sequences converge. So that's one example. When you're trying to show $T$ is closed, you get to assume sequences ${ p_n }$ and ${ Tp_n }$ converge to polynomials $x$ and $y$, respectively; you don't care how that may happen. Then you have to show that $y=Tx$; if you can establish that, then $T$ is closed. – Disintegrating By Parts Aug 16 '16 at 03:54