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Suppose $T$ is a bounded operator on a separable Hilbert space, and $Y$ and $Z$ two subspaces such that the restriction of $T$ to each of these subspaces has small norm. Does it follow that $T$ has small norm on the closed span of $Y\cup Z$? More precisely, if each of the restrictions has norm at most $\epsilon$, is the restriction to the closed span smaller than some $f(\epsilon)\to 0$?

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This is already false in $\mathbb{R}^2$: consider the projection $T(x_1, x_2)=(x_1, 0)$ and the subspaces $$ Y = \{(\epsilon t, t):t\in\mathbb{R}\}, \quad Z = \{(-\epsilon t, t):t\in\mathbb{R}\} $$ which span $\mathbb{R}^2$.

If $Y$ and $Z$ are orthogonal subspaces, then the answer is yes, since every element $x\in Y+Z$ can be written as $x=y+z$ where $y\in Y$, $z\in Z$, and consequently $\|x\|^2 = \|y\|^2+\|z\|^2$. It follows that $$ \|Tx\|\le \|Ty\|+\|Tz\|\le \epsilon(\|y\| + \|z\|)\le \epsilon\sqrt{2} \|x\| $$

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