Let $Y$ be a normed real vector space with the norm $||.||$, I am trying to see that this norm comes from an inner product on $Y$ if and only if for any $y,y'\in Y$ the function $Q(t)=||y+ty'||^2$ is quadratic on $\mathbb{R}.$
Now the way I am trying to do this is using the parallelogram law but I am getting nowhere. Suppose that the function is quadratic then for $x,y$ we will have $||x+ty||^2=a_2t^2 +a_1t+a_0$ and we will have that $||x+y||^2+||x-y||^2=2(a_0+a_2)=2(||x||^2+a_2)$, now the problem is that I can quite figure out what $a_2$ is, if I consider $||y+tx||^2=b_2t^2+b_1t+b_0$ I can see that we will have $b_1=a_1$ and $a_0+a_2=b_0+b_2$, just need to take $t=1$ and $t=-1$, where $b_0=||y||^2$ but I can't see why $a_2$ would be $||y||^2=b_0$.
Any help is aprecciated.
$\newcommand{nrm}[1]{\left\lVert {#1}\right\rVert}$You have that, for all $t\ne 0$, $\nrm{x+ty}^2=t^2\nrm{ \frac1tx+y}^2=b_2+ t b_1+t^2b_0$, and therefore $b_0=a_2$, which completes your proof.