Suppose that $ \{x_1,x_2,... \}$ is an orthonormal basis in a Hilbert space $H$ and that:
$$Y=\left \{y\in H: \sum_{n=1}^\infty\left(1+\frac{1}{n}\right)^2 |\langle y,x_n\rangle|^2\leq 1\right \}$$
Prove that $Y$ is a bounded, closed, convex set that has no element with greatest norm.
I have started the problem and observed that none of $\{x_1,...\}$ belong to $Y$. This means that any $y\in Y$ is of the form $y=c_1x_1+c_2x_2+...$ meaning that $y\in Y$ if and only if:
$$\sum_{n=1}^\infty \left(1+\frac{1}{n}\right)^2|c_n|^2\leq 1$$ From here I was able to confirm that any sequence of elements in $Y$ will converge to an element of $Y$ since the coefficients will converge. This verifies that $Y$ is closed. I am struggling with the rest.
To show that it is convex I tried taking $y_1,y_2\in Y$ and letting $t\in[0,1]$. Then I observed that coefficients of $ty_1+(1-t)y_2$ are of the form $ta_n+(1-t)b_n$ where $y_1=a_1x_1+a_2x_2+...$ and $y_2=b_1x_1+b_2x_2+...$. From this I looked at: $$\sum_{n=1}^\infty \left(1+\frac{1}{n}\right)|ta_n+(1-t)b_n|^2$$ Which works out (by the facts that $y_1,y_2\in Y$) that: $$\sum_{n=1}^\infty \left(1+\frac{1}{n}\right)|ta_n+(1-t)b_n|^2\leq1+t(1-t)\sum_{n=1}^\infty \left(1+\frac{1}{n}\right)|a_n||b_n|$$
I just need to show that $|a_n||b_n|=0$ for all $n$.
For bounded I'm not certain on the definition but don't I need to show that $\|y\|<r$ for every $y\in Y$ where $r$ is a positive real number?
For the final part to show that $Y$ has no element with greatest norm. I don't know where to start. I appreciate any help.
Looks like you have proved that $Y$ is bounded and closed space, but you don't know how to prove $Y$ is convex.
From Cauchy inequality $\langle y_1, y_2\rangle \leq \|y_1\|\cdot \|y_2\|\leq 1$.
We have $\|ty_1 - (1-t)y_2\|^2 = t^2\|y_1\|^2 + (1-t)^2\|y_2\|^2 + 2t(1-t)\langle y_1, y_2\rangle\leq t^2 + (1-t)^2 + 2t(1-t) = 1$.
Notice that we're not using norm and inner product defined in $H$. In this case, we're using the inner product $\langle y_1, y_2\rangle = \sum_{n = 1}^\infty(1+1/n)^2 a_nb_n$, and norm $\|y_1\| = \sum_{n=1}^\infty (1+1/n)^2 a_n^2$. In this case you should still need to prove this is a valid inner product.