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I have trouble with this problem

Let $X$ be the normed linear space of polynomials restricted to $[a, b]$ . For $P \in X$, define $\phi(P)$ to be the sum of the coefficients of $P$. Show that $\phi$ is linear. Is it continuous if $X$ has the topology induced by the maximum norm.

any hints!! thanks in advance.

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    $\begingroup$ $\phi$ is evaluation at $1$. whether or not $\phi$ is continuous depends on whether or not $1\in[a,b]$ $\endgroup$ – yoyo Mar 21 '14 at 2:06
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As mentioned by yoyo, the sum of the coefficients of $p$ is precisely $p(1)$.

If $1 \in [a,b]$, show directly that the linear functional $p \mapsto p(1)$ is continuous.

If $1 \notin [a,b]$, construct a polynomial with $p(1)=1$ but with $p$ uniformly close to 0 on $[a,b]$. The Weierstrass approximation theorem may be helpful.

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yes $\phi$ is linear !

that is because $X$ has the topology induced by its maximum norm !

this is a directly definition of polynomial norms!

see: http://mathworld.wolfram.com/PolynomialNorm.html

since, $P=\sum_{k=0}^{n}$$a_{k}z^{k}$

and, $\Arrowvert$$p$$\Arrowvert$$=\sum$$\arrowvert$${a_{k}}$${\arrowvert}$is bounded$\Longrightarrow$$\Arrowvert$$p$$\Arrowvert$$_{\infty}$$=max$$\arrowvert$${a_{k}}$${\arrowvert}$is bounded

so, it is continuous if it has the topology induced by its maximum norms!

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  • $\begingroup$ Assuming this is indeed the norm being used then consider the following. Let $p_n = \sum\limits_{k=0}^n z^k$ then $\phi(p_n) = \sum\limits_{k=0}^n 1 = n+1$. However, $||p_n||_\infty = max |1 | = 1$ for all $n \geq 0$. That is, $||\phi(p_n)|| = (n+1) ||p_n||$ so $\phi$ cannot be bounded. (If M is a proposed bound we need only look at $p_M$ to find a counterexample showing that M is too small.) $\endgroup$ – Cros Apr 10 '16 at 12:07

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