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Let us consider $L^p([0,1],\lambda)$ spaces, were $\lambda$ is simply the lebesgue measure. These are Banach spaces for $p\ge1$ (of course). It is well known that for $ 1\leq p < q \leq +\infty$ we have an inclusion (embedding) $$ T^q_p \ : \ L^q([0,1],\lambda) \rightarrow L^p([0,1], \lambda),$$ which is linear (thus also affine).

One thing I know about affine transformations of real $\mathbb{R}^{n}$ spaces is, that it can transform an ellipsoid into a sphere (and vice versa). This is why I wonder what can we say about this fact in the more abstract setting of $L^p$ spaces. Let as take $p=2$ and any $q>2$. Define $$S^q_2=\{f\in L^q:||f||_2=1\}.$$ Then $S_2^q$ considered in the $L^2$ space is a subset of the unit sphere. The question is

As $T_2^q(S_2^q)=S_2^q$, and $T_2^q$ is affine, does it hold true, that $S_{2}^q$ is "some kind" of an ellipsoid in the $L^q$ space?

If so, how is an appropriate notion of such ellipsoid defined? I would be glad for any insight.

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    $\begingroup$ I have problems with the generalization of the term "ellipsoid" to the infinite dimensional case. In the finite dimensional case the interior of the ellipsoid is a neighborhood of $0$. This no longer holds in the infinite dimensional case if the norms are not equivalent, as in your case with $p \not= q$, $\endgroup$ Commented Sep 26, 2020 at 10:42
  • $\begingroup$ Maybe it's a very thin ellipsoid? $\endgroup$
    – Ville Salo
    Commented Sep 26, 2020 at 11:14
  • $\begingroup$ A first easy setting is in Hilbert spaces. Then, it is essentially a matter of representation of self-adjoint bounded operators, and the situation is much the same as in finite dim. $\endgroup$ Commented Jun 12 at 8:43

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There is a result for bounded linear operators $T \colon \mathbb{H}_{1} \to \mathbb{H}_{2}$ between Hilbert spaces concerning the image of the unit ball. For $T$ being compact it states that the image of the unit ball is an ellipsoid. For the general case of $T$ it states that the image can be included into an ellipsoid (of course there is some statement about its structure). See, for example, Section 1.3 of Chapter V in R. Temam's "Infinite-Dimensional Dynamical Systems in Mechanics and Physics".

In a Hilbert space there is no problem to define an ellipsoid. You take axes $e_{k}$ (an orthonormal basis) and lengths $\alpha_{k} \geq 0$. Then $\mathcal{E} = \{ \sum_{k=1}^{\infty} (\frac{x_{k}}{\alpha_{k}})^{2} \leq 1 \}$ is your elliosoid. If you want to speak about ellipsoids in Banach spaces you should clarify what do you mean.

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