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Recall the parallelogram law:

$$\forall\,x,y\in X, \text{an inner product space, }\|x+y\|^2+\|x-y\|^2=2\|x\|^2+2\|y\|^2$$ This result can be used to check whether or not a norm has been induced by an inner product. Here are some examples:

  1. $\|\cdot\|_\infty$ on $\ell^\infty$ does not come from an inner product. To see this, take $x=(1,0,0,...)\in\ell^\infty$ and $y=(0,1,0,0,...)\in\ell^\infty$. But then, $\|x+y\|_\infty^2=\|x-y\|_\infty^2=\|x\|_\infty^2=\|y\|_\infty^2=1$, so that $\|x+y\|_\infty^2+\|x-y\|_\infty^2\neq2\|x\|_\infty^2+2\|y\|_\infty^2$. Thus, this norm is not induced by an inner product on $\ell^\infty$.
  2. $\|\cdot\|_1$ on $\ell^1$ does not come from an inner product. Using the same $x,y$ as for example 1., but regarding them here as elements in $\ell^1$, notice that $\|x+y\|_1^2=\|x-y\|_1^2=4$, but $\|x\|_1^2=\|y\|_1^2=1$. Then, putting it all together, we see that the parallelogram law isn't satisfied, so that this norm is not induced by an inner product on $\ell^1$.

These are some nice, easy to remember, examples which can be used with the parallelogram law to show that the respective norm is not induced by an inner product. Since the standard norm on $\ell^2$ is induced by an inner product, there isn't much to show with the Parallelogram law in the vein we did above.

But what about the Lebesgue measure, $L^p$, spaces for when $p\in[1,\infty)$? As well as in the case for $L^\infty$? Again, the norm on $L^2$ is induced by an inner product, but what about those $L^p$ spaces which are not? Which are those specific $L^p$ spaces whose norms are not induced by an inner product? And which simple, easy to remember, examples do you keep in your backpocket, ready to use in conjunction with the parallelogram law to show that?

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1 Answer 1

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I think $L^p$ are not induced by an inner product for $p \neq 2$. Indeed let $$f (x) = \mathbf 1_{[0,1]} (x)$$ and $$g (x) = \mathbf 1_{[1,2]} (x)$$ then $$\|f\|_p = \|g\|_p = 1$$

However $$\|f+g\|_p = \|f-g\|_p = 2^{\frac 1p}$$ then $$4 = 2\cdot 2^{\frac 2p} \Leftrightarrow p = 2$$

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  • $\begingroup$ That's actually really neat; this is definitely one to be kept in the back pocket because it encapsulates it for all of them for when $p\in[1,\infty)$. How about for when we have $L^\infty$? $\endgroup$
    – Jeremy Jeffrey James
    Commented May 8, 2018 at 19:02
  • $\begingroup$ The same example works for $L^\infty$. All what I wrote is correct with $p = \infty$. $\endgroup$
    – Kroki
    Commented May 8, 2018 at 19:04
  • $\begingroup$ Note that your example work in (almost) any measure space if you take any two disjoint finite measure sets $F,G$ and let $f=1_F/\mu(F)^{1/p}$, $g=1_G/\mu(G)^{1/p}$. $\endgroup$
    – Martin Argerami
    Commented May 8, 2018 at 19:21
  • $\begingroup$ Exactly. You can also probably add the fact that they are disjoint sets? $\endgroup$
    – Kroki
    Commented May 8, 2018 at 19:22

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