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Let $p \in \mathbb{C}[x]$ and $\mathbb{C}[x]$ is endowed with the norm $\|p\| = sup \{|p(t)|: t \in [0,1]\}$. I need to check if the functional $f(p) = p'(0)$ is continuous or not. I tend to think it's not. To prove it I tried to find some sequence in the unit ball which goes to unbounded set but I couldn't. I used the most simple ones like $2^{-n}(x+1)^n$ but it's image is bounded. What else can I do?

Invincible
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    By Weiestrass' theorem there is a sequence of polynomials converging uniformly to $\sqrt{x}$. –  Feb 06 '18 at 13:45
  • @treeleaf, if it is not continuous, we don't know how to extend it, so we don't know how our functional acts on $\sqrt{x}$ which gives us nothing. – Invincible Feb 06 '18 at 13:49
  • The functional doesn't need to act on $\sqrt{x}$ it acts on the sequence of polynomials. –  Feb 06 '18 at 13:50
  • @ treeleaf, okay, but how does it help? If we don't know how it acts on $\sqrt{x}$ we don't know that the sequence is unbounded, do we? – Invincible Feb 06 '18 at 13:52
  • Sit on it for a while. –  Feb 06 '18 at 13:55
  • @Vladislav: the functional $f$ has the domain $ \mathbb C[x]$. The function $h(x):= \sqrt{x}$ is not in the domain of $f$. Hence you can not use $h$ in the investigation of the continuity of$f$. – Fred Feb 06 '18 at 14:08

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Try $p_n(x)=(x-1)^n$. Then $||p_n||=1$ and $|f(p_n)|=|p_n'(0)|=n=n ||p_n||$

Fred
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