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As usual, $\mathbb R[x]$ denotes the vector space of polynomials in one variable with real coefficients. It is easy enough (and a good exercise for beginners) to prove that the function $$P\to\|P\|=\max_{x\in [0,1]} |P(x)|$$ of $\mathbb R[x]$ in $\mathbb R_+$ defines a norm on $\mathbb R[x]$ (then we can speak about continuity).

Prove that for all $x_0\in \mathbb R; x_0\gt 1$ the function $f_{x_0}$ of $\mathbb R[x]$ in $\mathbb R$ defined by $$f_{x_0}(P)= P(x_0)$$ is discontinuous in every point $P\in \mathbb R[x]$

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    $\begingroup$ It is false. If $x_0\in[0,1]$, then $f_{x_0}$ is continuous $\endgroup$ – sinbadh Mar 16 '16 at 14:51
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    $\begingroup$ This is not true. In fact $f_{x_0}$ is continuous if and only if $x_0\in[0,1]$. $\endgroup$ – David C. Ullrich Mar 16 '16 at 14:52
  • $\begingroup$ If you want a norm such that every $f_{x_0}$ is discontinuous you could for example define $||P||_1=\int_0^1|P(t)|\,dt$. $\endgroup$ – David C. Ullrich Mar 16 '16 at 14:53
  • $\begingroup$ I have edited the lapsus. Thank you very much. $\endgroup$ – Piquito Mar 16 '16 at 15:03
  • $\begingroup$ Pay attention please before deny. The question is true now. $\endgroup$ – Piquito Mar 16 '16 at 15:06
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Lemma: every continuous linear functional on a normed vector space is bounded, i.e. it has a finite operator norm.

Now the idea is that there are continuous functions which vanish on $[0,1]$ and are large at $x_0$, and therefore there are polynomials which are close to zero on $[0,1]$ but are large at $x_0$. This conclusion follows by Weierstrass' theorem. For details, let $f(x)=0$ on $[0,1]$, $1$ at $x_0$, and linear in between (the linearity is not important). Find a polynomial $p_n$ which is uniformly within $1/n$ of $f$. Then $P(p_n)=1$, but $\| p_n \| \leq 1/n$. Thus $\| P \|$, if $P$ were continuous, would need to be at least $n$, but this is impossible.

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  • $\begingroup$ Good. Whitout use of explicit theory, the sequence $\{P_n\}_{n\in \mathbb N}$ defined by $P_n(x)=(\frac xa)^n$; $1\lt a\lt x_0$ is such that $||P_n||=(\frac 1a)^n\to 0$ but $P_n(x_0)=(\frac {x_0}{a})^n\to \infty$ $\endgroup$ – Piquito Mar 16 '16 at 15:51
  • $\begingroup$ @Piquito Nice, that example does indeed work well. It also exposes the important fact that these examples require arbitrarily large degree. With a fixed maximum degree, the space is complete and homeomorphic to some $\mathbb{R}^n$, so none of these phenomena can occur. $\endgroup$ – Ian Mar 16 '16 at 16:04
  • $\begingroup$ In finite dimension there are not examples of discontinuous linear function. In other words, what you say yourself in your comment. Best regards. $\endgroup$ – Piquito Mar 16 '16 at 17:01

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