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.)
Best Answer
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.