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Let $P$ be the vector space of real valued polynomials over $R$. For any polynomial in $P$ set $p(t)=\sum_{k=0}^n a_kt^k $ and $\rVert p \lVert=\sum_{k=0}^n |a_k| $. I am being asked if the following linear maps $l:P \rightarrow R $ and $T: P \rightarrow P$: $$ l(p)=\int_0^1 p(t)dt $$ and $$(Tp)(t)=\int_0^tp(s)ds $$ are continous and determine their norm, $\rVert l \lVert $ and $\rVert T \lVert$. Following the definition, i will need to find a constant $M$ such that, for example in the second case, $\rVert Tp \lVert\leq M\rVert p \lVert, \forall p\in P$. At the other hand, following a theorem, a linear map on a finite- dimensional normed vector space is bounded and continous. I know that the vector space $P$ is not finite dimensional. So i am not sure if this theorem applies here. I am also wondering how the unit ball or sphere looks like at the vector space of the polynomials in order to determine $\rVert l \lVert$ and $\rVert T \lVert$.

Can somebody give a comment or any proposal ? Thanks.

user249018
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1 Answers1

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The space is not finite-dimensional, a basis for that is $\{1,x,x^{2},...\}$. $l$ is bounded because \begin{align*} |l(p)|\leq\int_{0}^{1}|p(t)|dt\leq\sum_{k=0}^{n}|a_{k}|\int_{0}^{1}t^{k}dt\leq\sum_{k=0}^{n}|a_{k}|=\|p\|, \end{align*} So the norm of $l$ satisfies $\|l\|\leq 1$, and actually it is $1$ because the constant polynomial $p=1$ is such that $l(p)=1$.

For the operator $T$, direct computation gives $(Tp)(t)=\displaystyle\sum_{k=0}^{n}\dfrac{a_{k}}{k+1}t^{k+1}$, so $\|Tp\|=\displaystyle\sum_{k=0}^{n}\dfrac{|a_{k}|}{k+1}\leq\sum_{k=0}^{n}|a_{k}|=\|p\|$, so $\|T\|\leq 1$, and again if we put the constant polynomial $p=1$, the norm is attained.

user284331
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  • Many thanks, its very clear. – user249018 Dec 01 '17 at 08:45
  • Yes, you are right. By the way, since $P$ is infinite-dimensional and $l$ is a linear functional from an infinite-dimensional space into $R$, what is the key point when proving that there exist an unbounded ( and thus not continous) linear functional on a infinite-dimensional normed vector space ? In other words, what is the space $P$ lacking so that $l$ is rather bounded and continous ? – user249018 Dec 01 '17 at 08:56
  • For any infinite-dimensional normed space, one can prove that there exists some unbounded linear operator, such a construction is delicate, so $P$ does not lack anything actually. – user284331 Dec 01 '17 at 17:13