# False statement that all norms on the direct sum of normed spaces are equivalent.

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

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)?

• 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 $p$. – s.harp Mar 23 '19 at 13:09
• 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 $||.||_q$? – vxnture Mar 23 '19 at 13:11
• $\|\cdot\|_p$ 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. – s.harp Mar 23 '19 at 13:28
• That makes so much more sense. Thank you! – vxnture Mar 23 '19 at 14:00

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.