Let $H$ be a Hilbert space, equipped with an inner product $(\cdot,\cdot)_1$ and norm $\|\cdot\|_1$ induced by it.

Let $(\cdot,\cdot)_2$ be other inner product on $H$ and $\|\cdot\|_2$ the norm induced by $(\cdot,\cdot)_2$.

Suppose there exists $a>0$ such that $\|x\|_2\leq a\|x\|_1$ for all $x\in H$. Is there any case on which this inequality implies that $(H,\|\cdot\|_2)$ is complete?

In other words, to prove that $(H,\|\cdot\|_2)$ is complete it's enough to show that there exists $a,b>0$ such that $\|x\|_2\leq a\|x\|_1$ and $\|x\|_1\leq b\|x\|_2$ for all $x\in H$ (that is, the norms are equivalents). I would like to know if there any case (maybe when we impose some condition on the inner product) on which $\|x\|_2\leq a\|x\|_1$ ensures the completeness.



Given that $H$ is complete in the $\lVert\,\cdot\,\rVert_1$ norm, and the estimate $\lVert x\rVert_2 \leqslant a\cdot \lVert x\rVert_1$ for all $x\in H$, by the open mapping theorem, $H$ is complete in the $\lVert\,\cdot\,\rVert_2$ norm if and only if you have an estimate $\lVert x\rVert_1 \leqslant b\cdot \lVert x\rVert_2$ for all $x\in H$.

If $H$ is finite-dimensional, it carries a unique Hausdorff topological vector space structure, and you are guaranteed such estimates. If $H$ is infinite-dimensional, an estimate $\lVert x\rVert_2 \leqslant a\cdot\lVert x\rVert_1$ for all $x\in H$ never guarantees the estimate in the other direction. If $(e_n)_{n\in\mathbb{N}}$ is an orthonormal system in $H$, denote its span by $T$, and for $x,y \in H$ consider

$$(x,y)_2 = (P x, Py)_1 + \sum_{n\in\mathbb{N}} 2^{-2n}(x,e_n)_1\cdot (e_n,y)_1,$$

where $P$ is the orthogonal projection onto $T^\perp$.

Then $\lVert x\rVert_2 \leqslant \lVert x\rVert_1$ for all $x$, but since $\lVert e_n\rVert_2 = 2^{-n}$, there is no estimate $\lVert x\rVert_1 \leqslant b\cdot \lVert x\rVert_2$ for all $x$, hence $H$ is not complete in the norm $\lVert\,\cdot\,\rVert_2$.

  • $\begingroup$ Nice. Another example: Let $H=L^2(\mathbb{R})$, and let $\langle f,g \rangle_2 = \int f(t)g(t) e^{-|t|} dt$. Then $f_n = 1_{[n,n+1]}$ is Cauchy with $\|\cdot\|_2$, but $\|f_n\|_1 = 1$ for all $n$. $\endgroup$ – copper.hat Feb 3 '14 at 17:47

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