# Is $f(t) = e^{X(t)}$ continuous when matrix $X(t)$ is continuous?

I think $$f(t) = e^{X(t)}$$ is continuous over $$t \in \mathbb{R}$$ when the complex matrix valued mapping $$t\mapsto X(t)$$ is continuous - regardless of whether $$X$$ is infinite or finite dimension. I tried to think of a proof, but without much success.

The hard part is the infinite-dimensional case, and I do not know what a proof could look like.

• The composed of two continuous maps is continuous. – Gustave Mar 10 at 11:51
• @Gustave: So you need to show $A \mapsto e^A$ is continuous. For marix $A$ or bounded operator $A$ it is clear. But the OP does not spell out his/her assumptions. Physicists talk about semigroups of unbounded operators, for example. – GEdgar Mar 10 at 12:34
• @GEdgar A semigroup by definition is continuous, we don't talk here about the $X(t)$ as operator but as one variable function. Cordially. – Gustave Mar 10 at 18:18