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I want to show that $\| \cdot \|_p$ is equivalent to $\| \cdot \|_q$ iff $p=q$ on $\mathbb{R}$.

I think the best way to do this is via a contradiction. Supposing that $\| \cdot \|_p \leq \alpha \| \cdot \|_q \leq \beta \| \cdot \|_p$, we have $$\left( \int_{\mathbb{R}} \left| f \right|^p d\mu \right)^{1/p} \leq \alpha \left( \int_{\mathbb{R}} \left| f \right|^q d\mu \right)^{1/q} \leq \beta \left( \int_{\mathbb{R}} \left| f \right|^p d\mu \right)^{1/p}.$$ I'm then thinking of applying Holder's inequality, but haven't been successful .

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HINT: we must show that if $q\neq p$ then these norms are not equivalent, that is doesnt exists $K\in(0,\infty)$ such that

$$K\|f\|_q\le\|f\|_p,\quad\forall f\in C(I,\Bbb K)$$

or equivalently

$$\frac{\|f\|_q}{\|f\|_p}\le K^{-1}$$

for $f\neq 0$. Choosing $f(x):=a^x$, $q=2$ and $p=1$ and $I:=[0,1]$ we have that it must be true that

$$\frac{\sqrt{\int_0^1 a^{2x}\mathrm dx}}{\int_0^1 a^x\mathrm dx}\le K^{-1},\quad\forall a>1$$

but the above is equivalent to say that

$$\frac{\sqrt{\int_0^1 a^{2x}\mathrm dx}}{\int_0^1 a^x\mathrm dx}=\frac{\sqrt{\frac{a^2-1}{2\ln a}}}{\frac{a-1}{\ln a}}=\sqrt{\frac{(a+1)\ln a}{(a-1)2}}\le K^{-1},\quad\forall a>1$$

what cannot be possible, hence $\|{\cdot}\|_2$ and $\|{\cdot}\|_1$ are not equivalent. We can generalize this result easily for any $q>p$ and any interval $[\alpha,\beta]$.

  • Dear @Masacroso, I would like to find a normed space with uncountable many norms pairwise not equivalent. You exhibited one. Do you think there is a simpler example? – rgm Apr 04 '18 at 13:58
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    @rgm I know some other space with non-equivalent norm, however it is not as easy like this one. The norms defined by $$|f|k:=\max{j=0,\ldots,k}|f^{(j)}|\infty$$ are not equivalent in any space of bounded smooth functions, by example on $C^\infty([0,1],\Bbb R)$. Here $f^{(j)}$ is the $j$-th derivative of $f$ and $|{\cdot}|\infty$ is the supremum norm. –  Apr 06 '18 at 19:49
  • Dear @Masacroso, I'd like one more help. I'm trying to generalize your result for any $q>p$. In my calculations, I need to use $(a^q-1)^p>(a^p-1)^q$. Is it necessary? Because I couldn't prove that. I fixed $I=[0,1]$ and set $a>e$, for convenience. – rgm Apr 20 '18 at 12:16
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    The generalization is very simple: for $f:[0,1]\to\Bbb R,, x\mapsto a^x$ for some $a>1$ we have for $p>q\ge 1$ that $$g(a):=\frac{|f|p}{|f|_q}=\frac{q^{1/q}}{p^{1/p}}\cdot\frac{(1-1/a^p)^{1/p}}{(1-1/a^q)^{1/q}}\cdot(\ln a)^{1/q-1/p}$$ Hence $$\lim{a\to\infty} g(a)=\infty$$ so the norms cannot be equivalents. –  Apr 20 '18 at 18:07