# Change of inner product on Hilbert space

Let $(\mathcal{V},\langle\cdot,\cdot\rangle_1)$ be a Hilbert space. If we change the inner product, can we then say anything about if that is a Hilbert space as well, i.e. when is $(\mathcal{V},\langle\cdot,\cdot\rangle_2)$ also a Hilbert space? Is it enough that $\langle\cdot,\cdot\rangle_2$ defines an inner product on $\mathcal{V}$?

• In finite dimension, yes. $\langle \cdot, \cdot \rangle_1$ and $\langle \cdot,\cdot \rangle_2$ induces norms $\|\cdot \|_1$ and $\|\cdot\|_2$, which will be equivalent. For infinite dimensional spaces I do not know an answer right now. – Ivo Terek Feb 20 '15 at 20:23

For example, let $H$ be any infinite dimensional Hilbert space,let $B$ be a Hamel basis of $H$ and consider the inner product on $H$ for which $B$ is an orthonormal basis. Then with that inner product $H$ is not complete.
• (+1) Thanks a lot for the answer. Do you know of any sufficient conditions that would do it? For example, what if $\langle\cdot,\cdot\rangle_2$ was coercive with respect to the $\|\cdot\|_1$-norm? That it is to say that for all $v\in\mathcal{V}$ it holds for some $\alpha>0$ that $\langle v,v\rangle_2 \geq \alpha \|v\|_1^2$. – Eff Feb 20 '15 at 20:40
• Under that condition you in fact that that $\Vert v\Vert_2\geq a\Vert v\Vert_1$, and then a Cauchy sequence wrt the new norm is also a Cauchy sequence wrt the old one, so yes that works. – Mariano Suárez-Álvarez Feb 20 '15 at 20:55
• Notice that the correct condition is «there is an $a>0$ such that for all $v\in V$ we have $\langle v,v\rangle_2\geq a\Vert v\Vert_1^2$» and not what you wrote: you have to be careful about where you place quantifiers in what you write! – Mariano Suárez-Álvarez Feb 20 '15 at 20:56