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The wikipedia entry for bounded operators shows that for the space $X$ of trigonometric polynomials on $[-\pi,\pi]$ with norm $$\lVert P\rVert = \int_{-\pi}^{\pi}\lvert P(x)\rvert \operatorname{d}\!x,$$ the operator $L: X\rightarrow X$ by $P\mapsto P^{\prime}$ is not a bounded operator.

It is not clear to me why there couldn't be something like a bounded linear operator $K$ on a larger space $X\subseteq Y$ such that the restriction of $K$ onto $X$ agrees with $L$. Is there an obvious uniqueness theorem I am missing?

I suspect that this can be proven by contradiction using the Riesz representation theorem, but I think this might be overkill and I am missing a comparatively trivial argument.

JessicaK
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  • I dont get your question. If the restriction of $K$ agrees with $L$, then $K$ can not be bounded as $L$ is not bounded. – user159517 Feb 04 '16 at 12:41
  • That is basically my question. I don't understand why that is the case and if it should be obvious or not. – JessicaK Feb 04 '16 at 12:43
  • $(e^{i n x})' = in e^{inx}$ so $||(e^{i n x})'|| = n ||e^{i n x}||$ i.e. your operator is not bounded (for the $||.||1$ norm) if on the $(e^{inx}){n \in \mathbb{N}}$ it is the derivative operator – reuns Feb 04 '16 at 12:45
  • If the operator $K$ identifies with the differential operator, and $K$ is bounded then the differential operator must also be bounded. Yes, it should be obvious. – Ranc Feb 04 '16 at 12:46

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The boundedness depends of the norm. $X\subseteq Y$ must be interpreted not only as set-theoretical inclusion, but that $X$ inherits the norm of $Y$.

Martín-Blas Pérez Pinilla
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  • and for any norm, the derivative operator on the $e^{inx}$ on $[-\pi;\pi]$ is unbounded – reuns Feb 04 '16 at 12:54
  • I think that is the big idea I was missing, but I will think about it for a little while before I accept an answer. Thank you. – JessicaK Feb 04 '16 at 13:00
  • @user1952009, take a finite-dimensional subspace... – Martín-Blas Pérez Pinilla Feb 04 '16 at 13:01
  • @Martín-Blas Pérez Pinilla : a linear operator on a subspace of $[-\pi;\pi] \to \mathbb{C}$ which is the derivative on the whole family $(e^{inx})_{n \in \mathbb{N}}$ cannot be bounded, whatever the norm considered – reuns Feb 04 '16 at 14:31
  • @user1952009, in a finite dimensional subspace any linear operator is bounded. See http://math.stackexchange.com/questions/112985/every-linear-mapping-on-a-finite-dimensional-space-is-continuous. The span of all the $x\mapsto e^{inx}$ is infinite dimensional. – Martín-Blas Pérez Pinilla Feb 04 '16 at 15:06
  • @user1952009, sorry I've misunderstand you. – Martín-Blas Pérez Pinilla Feb 04 '16 at 15:15
  • I think that's interesting, it justifies why on wikipedia we can read that the problem is that the derivative operator is unbounded no matter the norm considered. what saves us is that it is closed, so it keeps some of the properties of finite dimensional operators (it preserves the convergence of sequences for most (all?) norms) – reuns Feb 04 '16 at 15:20