On the sum of non-equivalent inner products Let $V$ be a complex vector space, equipped with two inner products $\langle \; , \; \rangle_1$ and $\langle \; , \; \rangle_2$, such that both $V_1 :=
 (V,\langle \; , \; \rangle_1)$ and $V_2 := (V,\langle \; , \; \rangle_2)$ are Hilbert spaces. Consider $\langle \; , \; \rangle_3 := \langle \; , \; \rangle_1 + \langle \; ,\; \rangle_2$ and the corresponding inner product space $V_3 := (V, \langle \; , \; \rangle_3)$. Now it is clear to me that there is no reason to assume that $V_3$ is again a Hilbert space in general, but I can't come up with a concrete example illustrating this phenomenon. 
Observe that this can only occur if $V$ is infinite dimesional and $\langle \; , \; \rangle_1$ and $\langle \; , \; \rangle_2$ are non-equivalent. In fact, to guarantee that $V_3$ is again a Hilbert space, it is sufficient to find some $c > 0$, such that for all $\forall v \in V$, we have $\langle v,v \rangle_2 \leq c \langle v,v\rangle_1$(In this case, $V_2$ doesn't even need to be a Hilbert space itself). All the examples I have in mind fulfill this condition, there doesn't seem to be an "obvious" example for when $V_3$ is not a Hilbert space. Does anyone have an idea ? 
 A: Take two cases.  The identity map $I_{1,2} \colon V_1 \to V_2$ has closed graph, or does not have closed graph.  In the first case, all three spaces $V_1, V_2, V_3$ have the same topology (using the closed graph theorem).  So your counterexample must be in the second case.
Write $\|x\|_1,\; \|x\|_2,\; \|x\|_3$ for the norms corresponding to the inner products.  Then
$$
\|x\|_3^2 = \|x\|_2^2 + \|x\|_2^2 .
\tag{$*$}
$$
Now suppose the identity $I_{1,2}$ does not have closed graph.  Then there is a sequence $z_n \in V$ such that $(z_n, z_n)$ in the graph converges to a point $(a,b)$ not in the graph.  That is, $\|z_n-a\|_1 \to 0$, $\|z_n-b\|_2 \to 0$, and $a \ne b$.  
I claim $V_3$ is not complete.  In fact, I claim that $z_n$ is a Cauchy sequence that does not converge.  
We know $z_n$ converges in $V_1$, so $z_n$ is Cauchy in $V_1$; that is $\|z_n-z_m\|_1 \to 0$ as $n,m \to \infty$.  Similarly $\|z_n-z_m\|_2 \to 0$.  And then by ($*$) we have $\|z_n-z_m\|_3 \to 0$.  Sequence $z_n$ is Cauchy in $V_3$.
Now I claim $z_n$ does not converge in $V_3$.  So assume it does, say $\|z_n-c\|_3 \to 0$.  Now by ($*$) we have $\|z_n-c\|_1 \le \|z_n-c\|_3$, so $z_n$ converges to $c$ in $V_1$.  Hence $c=a$.  Similarly $c=b$.  This contradiction shows $z_n$ does not converge in $V_3$.  So $V_3$ is not complete.
Remark Does this situation actually exist?  Is there an example of an infinite-dimensional vector space $V$ and two complete inner products on it so that the graph of the identity is not closed?  Yes.  But, in a sense, no.  We cannot explicitly write down such an example.  It requires the Axiom of Choice (or Hahn-Banach theorem, or other non-constructive principle beyond ZF set theory).
Take any infinite-dimensional Hilbert space $(V,\langle\cdot,\cdot\rangle_1)$ and any discontinuous linear bijection $I$ on it.  (A discontinuous linear bijection may be made easily from a Hamel basis, mapping each basis vector to a nonzero multiple of itself, but making sure the multiples used are unbounded.)  Then define $\langle x,y\rangle_2 := \langle Ix,Iy\rangle_1$.
