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I know that $l^2$ space is a Hilbert space. But for other $l^p$ spaces, where $p\geq1$, I have to show that they do not satisfy the parallelogram equality.

But, I can't find appropriate sequences that become counterexamples of the parallelogram equality. Could anyone suggest me some sequences?

Martin Sleziak
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Keith
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    Related: http://math.stackexchange.com/questions/997292/how-to-prove-that-lp-0-1-isnt-induced-by-an-inner-product-for-p-neq-2 – Martin Sleziak May 28 '15 at 11:23

1 Answers1

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Consider $\mathbf{x} = (1,0,0,0,\ldots)$ and $\mathbf{y} = (0,1,0,0,\ldots)$.

Then $\mathbf{x+y} = (1,1,0,0,\ldots)$ and $\mathbf{x-y} = (1,-1,0,0,\ldots)$.

$$\|\mathbf{x}+\mathbf{y}\|_p^2 + \|\mathbf{x}-\mathbf{y}\|_p^2 = 2\times(1+1)^\frac{2}{p} = 2^{1+\frac{2}{p}}$$

Meanwhile, $$2(\|\mathbf{x}\|_p^2+\|\mathbf{y}\|_p^2) = 2\times(1+1) = 4$$

If the parallelogram law holds, then $$2^{1+\frac{2}{p}} = 4 \implies 1+\frac{2}{p} = 2 \implies p = 2$$

Empiricist
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  • how do you know what $\Vert x +y\Vert^2_p$ is? It could be any inner product that induces the norm – Syd Dec 15 '20 at 14:10