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I've read somewhere that given two closed subspaces $V_1,V_2$ in topological vector space $X$, their algebraic span $V_1+V_2=\{x_1+x_2 |x_i \in V_i, i=1,2\}$ need not be closed. I always thought that such things only happen when we take infinite sums and would be interested in seeing example of such behaviour or proof that it's impossible. I'd also like to know if anything changes in concrete cases (e.g. Hilbert/Banach space instead of general TVS).

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There's an answer here. I'm going to reproduce it here to fill in a couple of details.

Let $$T : l^2 \rightarrow l^2 : (x_n) \mapsto \left(\frac{x_n}{n}\right).$$ Then $T$ is linear, and $$\|T(x_n)\|^2 = \left\|\left(\frac{x_n}{n}\right)\right\|^2 = \sum_{n=1}^\infty \frac{|x_n|^2}{n^2} \le \sum_{n=1}^\infty |x_n|^2 = \|(x_n)\|^2,$$ hence $T$ is bounded. Moreover, the range of $T$, the set $\lbrace (x_n) \in l^2 : (nx_n) \in l^2 \rbrace$ is not closed in $l^2$. For example, the sequence $(n^{-3/2}) \in l^2 \setminus T(l^2)$, but we can establish that it is in the closure of $T(l^2)$.

Consider a sequence of sequences $((n^{-3/2 - 1/k})_{n=1}^\infty)_{k=1}^\infty$ from $l^2$. Note that $(n \cdot n^{-3/2 - 1/k})^2 = n^{-1 - 2/k}$, the series of which converges, hence $(n^{-3/2 - 1/k})_{n=1}^\infty \in T(l^2)$. But then, $$\left\|n^{-3/2} - n^{-3/2 - 1/k}\right\|^2 = \sum_{n=2}^\infty n^{-3}\left(1 - n^{-1/k}\right)^2 \le \left(1 - 2^{-1/k}\right)^2\sum_{n=2}^\infty n^{-3} \rightarrow 0.$$ This proves $(n^{-3/2})$ is in the closure of $T(l^2)$.

Now we construct an actual example. Let $V = l^2 \oplus l^2$, a Hilbert space under the inner product $\langle (p, q), (r, s) \rangle_V = \langle p, r \rangle_{l^2} + \langle q, s \rangle_{l^2}$. Additionally, let $V_1$ be the graph of $T$, that is, $V_1 = \lbrace (x, Tx) : x \in l^2 \rbrace$ and $V_2 = l^2 \oplus \lbrace 0 \rbrace \subseteq V$. Note that $V_1$ is closed, since $T$ is continuous.

Notice that $(0, n^{-3/2}) \notin V_1 + V_2$. It is, however, in $\overline{V_1 + V_2}$, since $$(n^{-3/2-1/k}, T(n^{-3/2-1/k})) + (-n^{-3/2-1/k}, 0) \rightarrow (0, n^{-3/2})$$ as $k \rightarrow \infty$. This proves $V_1 + V_2$ is not closed.

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    $\begingroup$ Looks great, thanks. In calculation of norm of difference of two sequences you missed squares after the two parentheses. Of course it doesn't change the answer. I couldn't edit because it's less than 6 characters. $\endgroup$ – Blazej Jul 16 '15 at 11:59
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You can do it without any calculations: Let $T: X\to Y$ be a continuous linear map between two Banach spaces with dense range (e.g., $\ell^2\to\ell^2$, $(x_n)_n\mapsto (x_n/n)_n$ as in Theo's answer). Then take $Z=X\times Y$, $V_1=\lbrace (x,T(x)): x\in X\rbrace$ the graph of $T$ and $V_2= X\times \lbrace 0 \rbrace$. $V_1$ and $V_2$ are closed but their sum $V_1 + V_2 =X \times T(X)$ is dense. If $T$ is injective you additionally have $V_1 \cap V_2=\lbrace 0 \rbrace$.

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