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Let $(U, \|\cdot\|_U)$ be a Banach space, $(H_k, \|\cdot\|_k)_{k\in\mathbb{N}}$ be a sequence of Hilbert spaces and denote by

$$\tag{1}H:=\bigoplus_{k=1}^\infty H_k \equiv \left\{h=(h_k)\ \middle| \ h_k \in H_k, \,\forall k\in\mathbb{N} \quad \text{and} \quad \|h\|_H^2:=\sum_{k=1}^\infty\|h_k\|_k^2 < \infty \right\}$$

the $\ell^2$-direct sum of these spaces (which is known to be a Hilbert space itself).

Let further $\pi_k : H \rightarrow H_k$, $\pi_k((h_k)) := h_k$, be the projection of $H$ onto its $k^{\mathrm{th}}$-component.

Question: Is it true, then, that a (not necessarily linear) map $T : U \rightarrow H$ is $\textit{continuous}$ if its projections

$$\tag{2}T_k := \pi_k\circ T \quad \text{are continuous} \quad \text{for each } \ k\in\mathbb{N}?$$

Remark: If necessary, it may be assumed that each of the $H_k$ are finite-dimensional.

Any references, hints or proofs (or indeed counterexamples) that cover this are appreciated!

(This is not a homework question.)

fsp-b
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    This is not true and the reason is that the topology on $H$ is not the same as the product topology. In general, if $X_i$ is a collection of topological spaces, then one defines the product topology on $\prod_{i\in I}X_i$ and it is easily seen that a map $f:Y\to\prod_iX_i$ is continuous if and only if the compositions $\pi_j\circ f$ of $f$ with the projections are continuous. – Just dropped in Apr 27 '21 at 19:34
  • Also, when looking for a counter-example, it will be impossible to construct a linear, unbounded map $T:U\to H$, the reason behind this is that you require that $U$ is Banach (complete). We only know that unbounded operators with complete domains exist because of AC, but without it there is no guarantee for this. – Just dropped in Apr 27 '21 at 19:36
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    JustDroppedin, indeed, in the linear situation there is no counter-example because the closed graph theorem implies continuity here. – Ruy Apr 27 '21 at 20:59

1 Answers1

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Here is a counter example: take $U={\mathbb R}$ and $H_k={\mathbb R}$, for all $k$, so that $H=\ell ^2$.

Let $f:{\mathbb R}\to {\mathbb R}$ be the function whose graph is a triangle with vertices $(0,0)$, $(1,1)$, and $(2,0)$, hence vanishing identically on $(-\infty ,0]\cup [2,\infty )$, and consider the map $$ T:U\to H=\ell ^2, $$ given by $$ T(x) = \big (f(x), f(2x), f(3x), \ldots \big ). $$ Then clearly $\pi _k\circ T$ is continuous for all $k$. However $$ \|T(0)-T(1/n)\| = \|T(1/n)\| \geq $$$$ \geq \|\pi _n\big (T(1/n)\big ) )\| = |f(1)| = 1, $$ so $T(1/n)$ does not converge to $T(0)$, hence $T$ is not continuous at 0.

Ruy
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  • Nice example @Ruy. Note however that $T$ is not linear. This is OK for OP's purposes, but I wonder if we require $T$ to be linear. – J. De Ro Apr 27 '21 at 20:03
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    I think there is no linear counter-example because the hypotheses allow for an application of the closed graph theorem. – Ruy Apr 27 '21 at 20:57
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    @QuantumSpace I believe you're going to find the answer here interesting https://math.stackexchange.com/questions/2567528/any-examples-of-unbounded-linear-operators-between-ell-infty-and-ell-inft?fbclid=IwAR39qIOWvvWGdK3KwFJKoo9QVmCrmFHf0sh9wxYDxXuKpQj7XvsBL7JuMrU – Just dropped in Apr 27 '21 at 21:59
  • @JustDroppedIn Certainly. It goes in the list of things I should read soon :) – J. De Ro Apr 28 '21 at 09:02