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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
$begingroup$
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
New contributor
$endgroup$
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
New contributor
$endgroup$
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
New contributor
$endgroup$
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
real-analysis linear-algebra functional-analysis continuity matrix-exponential
New contributor
New contributor
edited 2 days ago
Rodrigo de Azevedo
13k41960
13k41960
New contributor
asked 2 days ago
Lucia GuzheimLucia Guzheim
134
134
New contributor
New contributor
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 |
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
add a comment |
0
active
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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