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I am currently working on the exercises in Conway's A Course in Functional Analysis and I think the following problem is not true. enter image description here

Here $\oplus_p X_k = \{(x_1, ..., x_n) \in \oplus_{k=1}^n X_k: (\sum_{k=1}^n ||x_k||_{X_k}^p)^{1/p} < \infty\} $. Obviously if each of the $X_i$ is finite-dimensional then so is any subspace of their direct sum, and any two norms on a finite-dimensional vector space are equivalent.

However, when $n=1$, and $X_1 = C([0,1], ||.||_1)$, where $||f||_1 = \int_{0}^{1}|f(x)|dx$, then $\oplus_p X_k = (C([0,1],||.||_1)$ and it is a standard result that the infinity norm $||.||_{\infty}$ is not equivalent to $||.||_1$ on $C([0,1])$.

Am I missing something? Or is there a correct formulation of the problem (that is a generalisation of the statement for finite-dimensional spaces)?

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    $\begingroup$ Yes, you are missing something. In the case $n=1$ then $\bigoplus_pX$ has norm $(\|x\|_{X_1}^{p})^{1/p}=\|x\|_{X_1}$, ie what you get doesn't depend on $p$. $\endgroup$ – s.harp Mar 23 '19 at 13:09
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    $\begingroup$ Ah, so is the problem to show that all norms on $\oplus_p X_k$ are equivalent to $||.||_p$, or that $||.||_p$ is equivalent to $||.||_q$? $\endgroup$ – vxnture Mar 23 '19 at 13:11
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    $\begingroup$ $\|\cdot\|_p$ is equivalent to $\|\cdot\|_q$, where these norms are the ones you have defined. Note that the problem "all norms on $\bigoplus_p X_k$ are equivalent" is not really well-defined. $\endgroup$ – s.harp Mar 23 '19 at 13:28
  • $\begingroup$ That makes so much more sense. Thank you! $\endgroup$ – vxnture Mar 23 '19 at 14:00
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It is probably meant that for all $1 \le p ,q \le \infty$ the norms $\|\cdot\|_p$ and $\|\cdot\|_q$ on the vector space $X_1 \oplus \cdots \oplus X_n$ given by $$\|x\|_p = \|(x_1, \ldots, x_n)\|_p = \left(\sum_{k=1}^n \|x_k\|_{X_k}^p\right)^{1/p}$$ $$\|x\|_q = \|(x_1, \ldots, x_n)\|_q = \left(\sum_{k=1}^n \|x_k\|_{X_k}^q\right)^{1/q}$$ are equivalent.

Assume $p \le q$. We have $\|x_k\|_{X_k} \le \|x\|_p$ so $$\frac{\|x\|_q^q}{\|x\|_p^q} = \sum_{k=1}^n\frac{\|x_k\|_{X_k}^q}{\|x\|_p^q} = \sum_{k=1}^n \left(\underbrace{\frac{\|x_k\|_{X_k}}{\|x\|_p}}_{\le 1}\right)^q \le \sum_{k=1}^n \left(\frac{\|x_k\|_{X_k}}{\|x\|_p}\right)^p = \sum_{k=1}^n\frac{\|x_k\|_{X_k}^p}{\|x\|_p^p} = \frac{\|x\|_p^p}{\|x\|_p^p} = 1$$ and therefore $\|x\|_q \le \|x\|_p$.

For the converse inequality use Hölder's inequality for conjugated exponents $\frac{q}{p}$ and $\frac1{1-\frac{q}p}$ to obtain $$\|x\|_p^p = \sum_{k=1}^n \|x_k\|_{X_k}^p \le \left(\sum_{k=1}^n \|x_k\|_{X_k}^{p\cdot\frac{q}{p}}\right)^{\frac{p}{q}}\left(\sum_{k=1}^n 1^{\frac1{1-\frac{q}p}}\right)^{1-\frac{p}{q}} = \left(\sum_{k=1}^n \|x_k\|_{X_k}^{q}\right)^{\frac{p}{q}} n^{1-\frac{p}q} = \|x\|_q^{p}n^{1-\frac{p}q}$$ It follows $\|x\|_p \le n^{\frac1p - \frac1q}\|x\|_q$. We conclude that the norms are equivalent.

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