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Let $P_2$ be the space of all polynomial of degree no more than 2 considered as functions on the interval $[0,2]$. For $p$ and $q$ in $P_2$, let $$\langle p,q\rangle=\int_0^2P(x) \overline {q(x)} \, dx$$ and let $\|\bullet\|_2$ be the norm associated with this inner product. Let $\|p\|_\infty =\sup\{|p(x)|, x\in [0,2]\}.$

For $p$ in $P_2$, let $\phi (p)=p(1)$.

I'm asked to show $\phi$ is continuous with respect to the norm $\|\bullet\|_\infty$ and also $\|\bullet\|_2$ on $P_2$

I know how to show a function is continuous but I've never done it with respect to a norm. I'd appreciate any help. Thanks

User69127
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    I changed $<p,q>$ to $\langle p,q\rangle$. I also changed $sup$ to $\sup$. The latter not only prevents italicization, but also (when in a "displayed" setting) affects positions of subscripts when one writes $\displaystyle\sup_{x\in S}$, and provides proper spacing in expressions like $a\sup b$. Also, notice that $||x||$ looks different from $|x|$. ${}\qquad{}$ – Michael Hardy Sep 07 '13 at 02:32
  • I could definitely use that math Jax sharpening. Thanks for the tip – User69127 Sep 07 '13 at 02:33
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    One way is to show that if $p_n \to p$ in one of the norms, then $\phi(p_n) \to \phi(p)$. $|\cdot|_\infty$ corresponds to uniform convergence of the polynomials, and $|\cdot|_2$ corresponds to $L^2$ convergence, slightly trickier. – Evan Sep 07 '13 at 03:08
  • @Evan what about finding a polynomial $q(x)$ in $P_2$ such that $\phi(p)=<p,q>$ for every $p$ in $P_2$ how do you think this can help? – User69127 Sep 07 '13 at 03:25
  • Ah yes, of course. Then it's easier to show that $\phi(p_n) \to \phi(p)$ by transferring to that inner product representation and using Cauchy Schwarz. – Evan Sep 07 '13 at 03:28

1 Answers1

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Recall what it means for a function to be continuous. We have $\phi:P_2 \to \mathbb{C}$. Continuity means that in order to control the output of $\phi$, it suffices to control the input of $\phi$. That is, $\phi$ is continuous at $x$ if for every neighborhood $V$ of $\phi(x)$, there is a neighborhood $U$ of $x$ such that $\phi(U) \subseteq V$. When we are working with normed spaces, this is equivalent to for every $\epsilon >0$ there is a $\delta>0$ such that $||x-y||<\delta$ implies $||\phi(x)-\phi(y)||<\epsilon$. Another equivalent formulation for normed spaces is that if $||x_n - x|| \to 0$, then $||\phi(x_n)-\phi(x)|| \to 0$.

What we have left off here is the specification of which norm. In our case, the norm on $\mathbb{C}$ is implicitly taken to be absolute value (i.e. the Euclidean norm). The point of the question is to change the norm on $P_2$. Explicitly, the question is asking you to show that

  1. If $||x_n-x||_2 \to 0$, then $\phi(x_n) \to \phi(x)$
  2. If $||x_n-x||_\infty \to 0$, then $\phi(x_n) \to \phi(x)$

I'll start. If $p_n \to p$ in the $||\cdot||_\infty$ norm this means that $\sup|p_n(x)-p(x)| \to 0$. In particular, we must have $|p_n(1)-p(1)| \to 0$, but this is exactly the statement that $\phi(p_n) \to \phi(p)$.

The other one is a little trickier. First can you see why $P_2$ is important? Try to come up with a non-polynomial example of where $f_n \to f$ in the $||\cdot||_2$ norm, but $f_n(1)$ does not tend to $f(1)$. Now why can't you create such an example with polynomials of degree at most 2?

Hint: Having bounded degree is important! Can you show that if $||p_n-p||_2 \to 0$, where $p_n(x) = a_nx^2 + b_nx + c, p(x) = ax^2+bx+c$, then you must have $a_n \to a, b_n \to b, c_n \to c$?

nullUser
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  • Can I also prove that $||\phi||$ is bounded and use the fact that boundedness implies continuous? – User69127 Sep 07 '13 at 17:10
  • @User69127 Yes you can because $\phi$ is a linear operator! – nullUser Sep 07 '13 at 17:48
  • yes. Showing that it's linear is easy...having trouble with the boundedness part :( – User69127 Sep 07 '13 at 19:31
  • I would try to use my hint and show it is continuous directly with sequences. Going through boundedness seems a little roundabout in this situation (though maybe you can think of a way to do it anyway). – nullUser Sep 07 '13 at 19:57
  • Yes I'll do it your way. I'm studying for some exit exams and these are some of the questions on the exams. I will ask the testing committee to see if they will accept this answer I more than likely think they want me to use boundedness. – User69127 Sep 07 '13 at 21:03