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


show $int_0^1 f(t)g(t) dt$ is a non-degenerate scalar productClassification theorem for vector spacesThe vector-valued distribution of compact supportDerivations on the spaces of continuous functions form an infinite dimensional vector space(generalized tangent space)Example of element of double dual that is not an evaluation mapClarification Question on meaning of 2-dimensional Rotation MatrixContinuous analogy of the matrix inversionDoes value of differentiation as infinity implies continuity?Prove that the spectral radius $rho(A)$ is a continuous function, where $A$ is a square matrix.Continuity of $textrmargmin$ set-valued mapping













0












$begingroup$


I think $f(t) = e^X(t)$ is continuous over $t in mathbbR$ when the complex matrix valued mapping $tmapsto X(t)$ 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.










share|cite|improve this question









New contributor




Lucia Guzheim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$







  • 3




    $begingroup$
    The composed of two continuous maps is continuous.
    $endgroup$
    – Gustave
    2 days ago











  • $begingroup$
    @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.
    $endgroup$
    – GEdgar
    2 days ago










  • $begingroup$
    @GEdgar A semigroup by definition is continuous, we don't talk here about the $X(t)$ as operator but as one variable function. Cordially.
    $endgroup$
    – Gustave
    2 days ago















0












$begingroup$


I think $f(t) = e^X(t)$ is continuous over $t in mathbbR$ when the complex matrix valued mapping $tmapsto X(t)$ 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.










share|cite|improve this question









New contributor




Lucia Guzheim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$







  • 3




    $begingroup$
    The composed of two continuous maps is continuous.
    $endgroup$
    – Gustave
    2 days ago











  • $begingroup$
    @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.
    $endgroup$
    – GEdgar
    2 days ago










  • $begingroup$
    @GEdgar A semigroup by definition is continuous, we don't talk here about the $X(t)$ as operator but as one variable function. Cordially.
    $endgroup$
    – Gustave
    2 days ago













0












0








0





$begingroup$


I think $f(t) = e^X(t)$ is continuous over $t in mathbbR$ when the complex matrix valued mapping $tmapsto X(t)$ 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.










share|cite|improve this question









New contributor




Lucia Guzheim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




I think $f(t) = e^X(t)$ is continuous over $t in mathbbR$ when the complex matrix valued mapping $tmapsto X(t)$ 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.







real-analysis linear-algebra functional-analysis continuity matrix-exponential






share|cite|improve this question









New contributor




Lucia Guzheim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




Lucia Guzheim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 2 days ago









Rodrigo de Azevedo

13k41960




13k41960






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Lucia Guzheim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 2 days ago









Lucia GuzheimLucia Guzheim

134




134




New contributor




Lucia Guzheim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor





Lucia Guzheim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Lucia Guzheim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







  • 3




    $begingroup$
    The composed of two continuous maps is continuous.
    $endgroup$
    – Gustave
    2 days ago











  • $begingroup$
    @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.
    $endgroup$
    – GEdgar
    2 days ago










  • $begingroup$
    @GEdgar A semigroup by definition is continuous, we don't talk here about the $X(t)$ as operator but as one variable function. Cordially.
    $endgroup$
    – Gustave
    2 days ago












  • 3




    $begingroup$
    The composed of two continuous maps is continuous.
    $endgroup$
    – Gustave
    2 days ago











  • $begingroup$
    @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.
    $endgroup$
    – GEdgar
    2 days ago










  • $begingroup$
    @GEdgar A semigroup by definition is continuous, we don't talk here about the $X(t)$ as operator but as one variable function. Cordially.
    $endgroup$
    – Gustave
    2 days ago







3




3




$begingroup$
The composed of two continuous maps is continuous.
$endgroup$
– Gustave
2 days ago





$begingroup$
The composed of two continuous maps is continuous.
$endgroup$
– Gustave
2 days ago













$begingroup$
@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.
$endgroup$
– GEdgar
2 days ago




$begingroup$
@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.
$endgroup$
– GEdgar
2 days ago












$begingroup$
@GEdgar A semigroup by definition is continuous, we don't talk here about the $X(t)$ as operator but as one variable function. Cordially.
$endgroup$
– Gustave
2 days ago




$begingroup$
@GEdgar A semigroup by definition is continuous, we don't talk here about the $X(t)$ as operator but as one variable function. Cordially.
$endgroup$
– Gustave
2 days ago










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