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

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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

edited 2 days ago

Rodrigo de Azevedo

13k41960

asked 2 days ago

Lucia GuzheimLucia Guzheim

134

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
  

  
  
  
  

add a comment | 

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.

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

share|cite|improve this question

edited 2 days ago

Rodrigo de Azevedo

13k41960

asked 2 days ago

Lucia GuzheimLucia Guzheim

134

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
  

  
  
  
  

add a comment | 

$begingroup$

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.

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

share|cite|improve this question

edited 2 days ago

Rodrigo de Azevedo

13k41960

asked 2 days ago

Lucia GuzheimLucia Guzheim

134

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.

$endgroup$

share|cite|improve this question

edited 2 days ago

Rodrigo de Azevedo

13k41960

asked 2 days ago

Lucia GuzheimLucia Guzheim

134

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

edited 2 days ago

Rodrigo de Azevedo

13k41960

asked 2 days ago

Lucia GuzheimLucia Guzheim

134

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.

edited 2 days ago

Rodrigo de Azevedo

13k41960

edited 2 days ago

Rodrigo de Azevedo

13k41960

asked 2 days ago

Lucia GuzheimLucia Guzheim

134

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.

asked 2 days ago

Lucia GuzheimLucia Guzheim

134