Show that the intersection of $\overline{\text{span }\{e_{2n}\}}$ and $\overline{\text{span }\{e_{2n}+\frac{1}{n+1}e_{2n+1}\}}$ is zero.

Question

Let $(e_n)_{n\in\mathbb{N}}$ be a complete orthonormal system in a Hilbert space $H$. Let $x_n=e_{2n}$ and $y_n=x_n + \frac{1}{n+1}e_{2n+1}$. Let $F=\overline{\text{span }\{x_n, n\in\mathbb{N}\}}$ and $G=\overline{\text{span }\{y_n, n\in\mathbb{N}\}}$.

I need to show that any element in $F+G=\{z\in H \mid z=x+y,\quad x\in F, y\in G\}$ can be written as an element from $F$ and an element from $G$ in only one way.

When the intersection of $F$ and $G$ is the zero vector space, I can prove the statement above, but I do not know how to show $F\cap G=\{0\}$.

This means that no element in $F$ is in $G$ and vice versa.

What bothers me is that my intuition makes me feel like the intersection is not zero since $y_n = x_n + \frac{1}{n+1}e_{2n+1}$ converges to $x_n$ as $n$ becomes infinitely large, since the norm of $e_k$ is always 1. Therefore this would make it seem that $y_n=x_n\in G$ for $n\to\infty$, so $F\cap G\supset\{0,x_n\}\neq\{0\}$.

Any ideas? I know that if $z=(x+y)\in H$, then $(x+y)=\Sigma \langle (x+y)\mid e_k\rangle e_k$, since $(e_k)$ is a complete orthornormal system in $H$.

Answer 1

The error in your intuition lies in the statement "$y_n = x_n + \frac{1}{n+1}e_{2n+1}$ converges to $x_n$ as $n$ becomes infinitely large". If you are letting $n$ tend to infinity, then any individual $x_n$ does not have meaning in that context.

Here's the same fallacy in a more elementary situation: "the function $y=x+1/x$ converges to $x$ as $x$ becomes infinitely large, so $y=x$ for $x\to\infty$".

Answer 2

It is demonstrated that $F$ and $G$ intersect trivially, as prompted in the title.

$F,G\subset H$ are closed subspaces, hence Hilbert spaces in their own. An ONB in $F$ is $\{e_{2k}, k=0,1,2,\dots\}$, and in $G$ consider the ONB $\big\{f_{2k}=\gamma_k\big(e_{2k}+\frac1{k+1}e_{2k+1}\big), k=0,1,\dots\big\}$ with normalisation factors $\,\gamma_k = \frac{k+1}{\sqrt{(k+1)^2+1}}$.

Let $z$ like "zero" be in $F\cap G$, and let $$\sum_{k=0}^\infty\alpha_k e_{2k}\,=\; z \;=\,\sum_{k=0}^\infty\beta_k f_{2k}$$ be the Fourier expansions of $z$ in $F$ and $G$, respectively.
Denote the partial sums by $\,a_n=\sum_{k=0}^n\alpha_k e_{2k}\,$ and $\,b_n=\sum_{k=0}^n\beta_k f_{2k}$, they converge in norm to $z$.

Fix $\varepsilon>0$. Then choose $N$ such that both $\,\|b_n-z\|,\|z-a_n\|< \frac{1}{2}\sqrt\varepsilon\,$ for all $n\geqslant N$. Then $\|b_n-a_n\|<\sqrt\varepsilon\,$ by the triangle inequality, and $$\begin{align}\varepsilon \;>\;\|b_n-a_n\|^2 & \;=\; \Big\|\sum\nolimits_{k=0}^n\big(\beta_k\gamma_k\, e_{2k} + \beta_k\frac{\gamma_k}{k+1}e_{2k+1} - \alpha_k e_{2k}\big)\Big\|^2 \\[1ex] & \;=\;\sum_{k=0}^n\Big(|\beta_k\gamma_k- \alpha_k|^2 + |\beta_k|^2\frac{\gamma_k^2}{(k+1)^2}\Big)\end{align}$$ as $\{e_0,e_1,\dots,e_{2k},e_{2k+1},\dots\}$ is an ONB in $H$.
Now in the last sum each summand $<\varepsilon$ by necessity, and $\varepsilon>0$ was at choice. Hence $\beta_k=0\,$ for all $k$, which means $z=0$ and $F\cap G=\{0\}\,. \qquad\square$