$C^{infty}(mathbb{R})$ as a Fréchet space












1














I think I'm misunderstanding what is being asked of me in the following question:




The space $C^{infty}(mathbb{R})$ has a Fréchet space topology with respect to which $f_n rightarrow f$ iff $f^{(k)}_n rightarrow f^{(k)}$ uniformly on all compact subsets ($k$, non-negative, is the order of derivative).




To show the backwards implication I need to show that convergence ($f^{(k)}_n rightarrow f^{(k)}$ uniformly on all compact subsets) give me seminorms, there are countably many such semi norms, and these seminorms give me a complete Hausdorff TVS. My book gives me some facts to help me in this direction.



I'm confused by the forward implication though - if endow the space of $C^{infty}(mathbb{R})$ with a topology given by countable seminorms which makes $C^{infty}(mathbb{R})$ a complete Hausdorff TVS -- how do I know this implies $f^{(k)}_n rightarrow f^{(k)}$ uniformly on all compact subsets? What if the seminorms are really strange and have nothing to do with the derivatives of $f$ over compact subsets?






real-analysis general-topology functional-analysis topological-vector-spaces







share|cite|improve this question




















  • 1




    The seminirms are $p_{n,k}(f):= sup_{[-k,k]} |f^{(n)}|$. The key is that any compact set is contained in some interval $[-k,k]$
    – Tito Eliatron
    Nov 30 at 16:58












  • I see that, but that's for a particular choice of seminorms. Why should it be true for any arbitrary countable family of seminorms though?
    – yoshi
    Nov 30 at 17:21






  • 1




    For fixed k,n,m, The set ${p_{n,k}(f,0)<m}$ is open, so for your Family of seminorms, you can put a p- Bail inside.
    – Tito Eliatron
    Nov 30 at 17:31










  • Can you elaborate on this statement, what is a p-ball and why is it suffient for one p-ball to be inside?
    – yoshi
    Dec 1 at 15:20
















1














I think I'm misunderstanding what is being asked of me in the following question:




The space $C^{infty}(mathbb{R})$ has a Fréchet space topology with respect to which $f_n rightarrow f$ iff $f^{(k)}_n rightarrow f^{(k)}$ uniformly on all compact subsets ($k$, non-negative, is the order of derivative).




To show the backwards implication I need to show that convergence ($f^{(k)}_n rightarrow f^{(k)}$ uniformly on all compact subsets) give me seminorms, there are countably many such semi norms, and these seminorms give me a complete Hausdorff TVS. My book gives me some facts to help me in this direction.



I'm confused by the forward implication though - if endow the space of $C^{infty}(mathbb{R})$ with a topology given by countable seminorms which makes $C^{infty}(mathbb{R})$ a complete Hausdorff TVS -- how do I know this implies $f^{(k)}_n rightarrow f^{(k)}$ uniformly on all compact subsets? What if the seminorms are really strange and have nothing to do with the derivatives of $f$ over compact subsets?






real-analysis general-topology functional-analysis topological-vector-spaces







share|cite|improve this question




















  • 1




    The seminirms are $p_{n,k}(f):= sup_{[-k,k]} |f^{(n)}|$. The key is that any compact set is contained in some interval $[-k,k]$
    – Tito Eliatron
    Nov 30 at 16:58












  • I see that, but that's for a particular choice of seminorms. Why should it be true for any arbitrary countable family of seminorms though?
    – yoshi
    Nov 30 at 17:21






  • 1




    For fixed k,n,m, The set ${p_{n,k}(f,0)<m}$ is open, so for your Family of seminorms, you can put a p- Bail inside.
    – Tito Eliatron
    Nov 30 at 17:31










  • Can you elaborate on this statement, what is a p-ball and why is it suffient for one p-ball to be inside?
    – yoshi
    Dec 1 at 15:20














1












1








1







I think I'm misunderstanding what is being asked of me in the following question:




The space $C^{infty}(mathbb{R})$ has a Fréchet space topology with respect to which $f_n rightarrow f$ iff $f^{(k)}_n rightarrow f^{(k)}$ uniformly on all compact subsets ($k$, non-negative, is the order of derivative).




To show the backwards implication I need to show that convergence ($f^{(k)}_n rightarrow f^{(k)}$ uniformly on all compact subsets) give me seminorms, there are countably many such semi norms, and these seminorms give me a complete Hausdorff TVS. My book gives me some facts to help me in this direction.



I'm confused by the forward implication though - if endow the space of $C^{infty}(mathbb{R})$ with a topology given by countable seminorms which makes $C^{infty}(mathbb{R})$ a complete Hausdorff TVS -- how do I know this implies $f^{(k)}_n rightarrow f^{(k)}$ uniformly on all compact subsets? What if the seminorms are really strange and have nothing to do with the derivatives of $f$ over compact subsets?






real-analysis general-topology functional-analysis topological-vector-spaces







share|cite|improve this question















I think I'm misunderstanding what is being asked of me in the following question:




The space $C^{infty}(mathbb{R})$ has a Fréchet space topology with respect to which $f_n rightarrow f$ iff $f^{(k)}_n rightarrow f^{(k)}$ uniformly on all compact subsets ($k$, non-negative, is the order of derivative).




To show the backwards implication I need to show that convergence ($f^{(k)}_n rightarrow f^{(k)}$ uniformly on all compact subsets) give me seminorms, there are countably many such semi norms, and these seminorms give me a complete Hausdorff TVS. My book gives me some facts to help me in this direction.



I'm confused by the forward implication though - if endow the space of $C^{infty}(mathbb{R})$ with a topology given by countable seminorms which makes $C^{infty}(mathbb{R})$ a complete Hausdorff TVS -- how do I know this implies $f^{(k)}_n rightarrow f^{(k)}$ uniformly on all compact subsets? What if the seminorms are really strange and have nothing to do with the derivatives of $f$ over compact subsets?






real-analysis general-topology functional-analysis topological-vector-spaces




real-analysis general-topology functional-analysis topological-vector-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 30 at 17:37









Bernard

118k638111




118k638111










asked Nov 30 at 16:27









yoshi

1,170817




1,170817








  • 1




    The seminirms are $p_{n,k}(f):= sup_{[-k,k]} |f^{(n)}|$. The key is that any compact set is contained in some interval $[-k,k]$
    – Tito Eliatron
    Nov 30 at 16:58












  • I see that, but that's for a particular choice of seminorms. Why should it be true for any arbitrary countable family of seminorms though?
    – yoshi
    Nov 30 at 17:21






  • 1




    For fixed k,n,m, The set ${p_{n,k}(f,0)<m}$ is open, so for your Family of seminorms, you can put a p- Bail inside.
    – Tito Eliatron
    Nov 30 at 17:31










  • Can you elaborate on this statement, what is a p-ball and why is it suffient for one p-ball to be inside?
    – yoshi
    Dec 1 at 15:20














  • 1




    The seminirms are $p_{n,k}(f):= sup_{[-k,k]} |f^{(n)}|$. The key is that any compact set is contained in some interval $[-k,k]$
    – Tito Eliatron
    Nov 30 at 16:58












  • I see that, but that's for a particular choice of seminorms. Why should it be true for any arbitrary countable family of seminorms though?
    – yoshi
    Nov 30 at 17:21






  • 1




    For fixed k,n,m, The set ${p_{n,k}(f,0)<m}$ is open, so for your Family of seminorms, you can put a p- Bail inside.
    – Tito Eliatron
    Nov 30 at 17:31










  • Can you elaborate on this statement, what is a p-ball and why is it suffient for one p-ball to be inside?
    – yoshi
    Dec 1 at 15:20








1




1




The seminirms are $p_{n,k}(f):= sup_{[-k,k]} |f^{(n)}|$. The key is that any compact set is contained in some interval $[-k,k]$
– Tito Eliatron
Nov 30 at 16:58






The seminirms are $p_{n,k}(f):= sup_{[-k,k]} |f^{(n)}|$. The key is that any compact set is contained in some interval $[-k,k]$
– Tito Eliatron
Nov 30 at 16:58














I see that, but that's for a particular choice of seminorms. Why should it be true for any arbitrary countable family of seminorms though?
– yoshi
Nov 30 at 17:21




I see that, but that's for a particular choice of seminorms. Why should it be true for any arbitrary countable family of seminorms though?
– yoshi
Nov 30 at 17:21




1




1




For fixed k,n,m, The set ${p_{n,k}(f,0)<m}$ is open, so for your Family of seminorms, you can put a p- Bail inside.
– Tito Eliatron
Nov 30 at 17:31




For fixed k,n,m, The set ${p_{n,k}(f,0)<m}$ is open, so for your Family of seminorms, you can put a p- Bail inside.
– Tito Eliatron
Nov 30 at 17:31












Can you elaborate on this statement, what is a p-ball and why is it suffient for one p-ball to be inside?
– yoshi
Dec 1 at 15:20




Can you elaborate on this statement, what is a p-ball and why is it suffient for one p-ball to be inside?
– yoshi
Dec 1 at 15:20










1 Answer
1






active

oldest

votes


















1














The question doesn't exclude that there are different topologies on $C^infty(mathbb{R})$ such that $C^infty(mathbb{R})$ is a Frechet space with this topology. However, since the open mapping theorem is also true for Frechet spaces, we cannot have a finer (or weaker) topology (as the usual one) on $C^infty(mathbb{R})$ in which $C^infty(mathbb{R})$ is also a Frechet space and $f^{(k)}_n rightarrow f^{(k)}$ uniformly on compact sets for each $k in mathbb{N}$.



One can proof that $C^infty(mathbb{R})$ has dimension $2^{aleph_0}$ in the sense of Hamel basis. Thus you can take e.g. the Banach-space $L^1(mathbb{R})$ and transport the norm to $C^infty(mathbb{R})$. With these norm $C^infty(mathbb{R})$ is a Banach space. Since the topology of local convergence of all derivates is not normable, this weird topology is different than the usual one.






share|cite|improve this answer























  • why does "no finer topology" follow from the open mapping theorem?
    – yoshi
    Dec 1 at 15:22








  • 1




    If $tau_1$, $tau_2$ are topologies on $C^infty(mathbb{R})$ with $tau_1 subset tau_2$, i.e. $tau_2$ is finer than $tau_1$, then $mathrm{id} colon (C^infty(mathbb{R}),tau_2) rightarrow (C^infty(mathbb{R}),tau_1)$ the map is continuous. If both topologies are Frechet, then the open mapping theorem already implies that this map is open, i.e. the inverse map $mathrm{id} colon (C^infty(mathbb{R}),tau_1) rightarrow (C^infty(mathbb{R}),tau_2)$ is continuous and thus $tau_1 = tau_2$.
    – p4sch
    Dec 1 at 21:27












  • I like this, but my only problem is that we haven't covered the open mapping theorem for Frechet spaces. Is there a way to adapt Tito Aliatron's hints into this? I'm looking through the proof of the open mapping theorem for Frechet spaces to see if I can do it too.
    – yoshi
    Dec 1 at 21:44








  • 1




    I don't think that you can prove this much easier. It is a feature coming from the completeness of the space. You may have a look in Rudin's book on functional analysis. There you can find a proof of the open mapping theorem - that is Theorem 2.11. In fact, the idea of the proof is the same as in the case of banach spaces.
    – p4sch
    Dec 2 at 11:03











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1 Answer
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active

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1














The question doesn't exclude that there are different topologies on $C^infty(mathbb{R})$ such that $C^infty(mathbb{R})$ is a Frechet space with this topology. However, since the open mapping theorem is also true for Frechet spaces, we cannot have a finer (or weaker) topology (as the usual one) on $C^infty(mathbb{R})$ in which $C^infty(mathbb{R})$ is also a Frechet space and $f^{(k)}_n rightarrow f^{(k)}$ uniformly on compact sets for each $k in mathbb{N}$.



One can proof that $C^infty(mathbb{R})$ has dimension $2^{aleph_0}$ in the sense of Hamel basis. Thus you can take e.g. the Banach-space $L^1(mathbb{R})$ and transport the norm to $C^infty(mathbb{R})$. With these norm $C^infty(mathbb{R})$ is a Banach space. Since the topology of local convergence of all derivates is not normable, this weird topology is different than the usual one.






share|cite|improve this answer























  • why does "no finer topology" follow from the open mapping theorem?
    – yoshi
    Dec 1 at 15:22








  • 1




    If $tau_1$, $tau_2$ are topologies on $C^infty(mathbb{R})$ with $tau_1 subset tau_2$, i.e. $tau_2$ is finer than $tau_1$, then $mathrm{id} colon (C^infty(mathbb{R}),tau_2) rightarrow (C^infty(mathbb{R}),tau_1)$ the map is continuous. If both topologies are Frechet, then the open mapping theorem already implies that this map is open, i.e. the inverse map $mathrm{id} colon (C^infty(mathbb{R}),tau_1) rightarrow (C^infty(mathbb{R}),tau_2)$ is continuous and thus $tau_1 = tau_2$.
    – p4sch
    Dec 1 at 21:27












  • I like this, but my only problem is that we haven't covered the open mapping theorem for Frechet spaces. Is there a way to adapt Tito Aliatron's hints into this? I'm looking through the proof of the open mapping theorem for Frechet spaces to see if I can do it too.
    – yoshi
    Dec 1 at 21:44








  • 1




    I don't think that you can prove this much easier. It is a feature coming from the completeness of the space. You may have a look in Rudin's book on functional analysis. There you can find a proof of the open mapping theorem - that is Theorem 2.11. In fact, the idea of the proof is the same as in the case of banach spaces.
    – p4sch
    Dec 2 at 11:03
















1














The question doesn't exclude that there are different topologies on $C^infty(mathbb{R})$ such that $C^infty(mathbb{R})$ is a Frechet space with this topology. However, since the open mapping theorem is also true for Frechet spaces, we cannot have a finer (or weaker) topology (as the usual one) on $C^infty(mathbb{R})$ in which $C^infty(mathbb{R})$ is also a Frechet space and $f^{(k)}_n rightarrow f^{(k)}$ uniformly on compact sets for each $k in mathbb{N}$.



One can proof that $C^infty(mathbb{R})$ has dimension $2^{aleph_0}$ in the sense of Hamel basis. Thus you can take e.g. the Banach-space $L^1(mathbb{R})$ and transport the norm to $C^infty(mathbb{R})$. With these norm $C^infty(mathbb{R})$ is a Banach space. Since the topology of local convergence of all derivates is not normable, this weird topology is different than the usual one.






share|cite|improve this answer























  • why does "no finer topology" follow from the open mapping theorem?
    – yoshi
    Dec 1 at 15:22








  • 1




    If $tau_1$, $tau_2$ are topologies on $C^infty(mathbb{R})$ with $tau_1 subset tau_2$, i.e. $tau_2$ is finer than $tau_1$, then $mathrm{id} colon (C^infty(mathbb{R}),tau_2) rightarrow (C^infty(mathbb{R}),tau_1)$ the map is continuous. If both topologies are Frechet, then the open mapping theorem already implies that this map is open, i.e. the inverse map $mathrm{id} colon (C^infty(mathbb{R}),tau_1) rightarrow (C^infty(mathbb{R}),tau_2)$ is continuous and thus $tau_1 = tau_2$.
    – p4sch
    Dec 1 at 21:27












  • I like this, but my only problem is that we haven't covered the open mapping theorem for Frechet spaces. Is there a way to adapt Tito Aliatron's hints into this? I'm looking through the proof of the open mapping theorem for Frechet spaces to see if I can do it too.
    – yoshi
    Dec 1 at 21:44








  • 1




    I don't think that you can prove this much easier. It is a feature coming from the completeness of the space. You may have a look in Rudin's book on functional analysis. There you can find a proof of the open mapping theorem - that is Theorem 2.11. In fact, the idea of the proof is the same as in the case of banach spaces.
    – p4sch
    Dec 2 at 11:03














1












1








1






The question doesn't exclude that there are different topologies on $C^infty(mathbb{R})$ such that $C^infty(mathbb{R})$ is a Frechet space with this topology. However, since the open mapping theorem is also true for Frechet spaces, we cannot have a finer (or weaker) topology (as the usual one) on $C^infty(mathbb{R})$ in which $C^infty(mathbb{R})$ is also a Frechet space and $f^{(k)}_n rightarrow f^{(k)}$ uniformly on compact sets for each $k in mathbb{N}$.



One can proof that $C^infty(mathbb{R})$ has dimension $2^{aleph_0}$ in the sense of Hamel basis. Thus you can take e.g. the Banach-space $L^1(mathbb{R})$ and transport the norm to $C^infty(mathbb{R})$. With these norm $C^infty(mathbb{R})$ is a Banach space. Since the topology of local convergence of all derivates is not normable, this weird topology is different than the usual one.






share|cite|improve this answer














The question doesn't exclude that there are different topologies on $C^infty(mathbb{R})$ such that $C^infty(mathbb{R})$ is a Frechet space with this topology. However, since the open mapping theorem is also true for Frechet spaces, we cannot have a finer (or weaker) topology (as the usual one) on $C^infty(mathbb{R})$ in which $C^infty(mathbb{R})$ is also a Frechet space and $f^{(k)}_n rightarrow f^{(k)}$ uniformly on compact sets for each $k in mathbb{N}$.



One can proof that $C^infty(mathbb{R})$ has dimension $2^{aleph_0}$ in the sense of Hamel basis. Thus you can take e.g. the Banach-space $L^1(mathbb{R})$ and transport the norm to $C^infty(mathbb{R})$. With these norm $C^infty(mathbb{R})$ is a Banach space. Since the topology of local convergence of all derivates is not normable, this weird topology is different than the usual one.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 1 at 7:17

























answered Nov 30 at 20:33









p4sch

4,800217




4,800217












  • why does "no finer topology" follow from the open mapping theorem?
    – yoshi
    Dec 1 at 15:22








  • 1




    If $tau_1$, $tau_2$ are topologies on $C^infty(mathbb{R})$ with $tau_1 subset tau_2$, i.e. $tau_2$ is finer than $tau_1$, then $mathrm{id} colon (C^infty(mathbb{R}),tau_2) rightarrow (C^infty(mathbb{R}),tau_1)$ the map is continuous. If both topologies are Frechet, then the open mapping theorem already implies that this map is open, i.e. the inverse map $mathrm{id} colon (C^infty(mathbb{R}),tau_1) rightarrow (C^infty(mathbb{R}),tau_2)$ is continuous and thus $tau_1 = tau_2$.
    – p4sch
    Dec 1 at 21:27












  • I like this, but my only problem is that we haven't covered the open mapping theorem for Frechet spaces. Is there a way to adapt Tito Aliatron's hints into this? I'm looking through the proof of the open mapping theorem for Frechet spaces to see if I can do it too.
    – yoshi
    Dec 1 at 21:44








  • 1




    I don't think that you can prove this much easier. It is a feature coming from the completeness of the space. You may have a look in Rudin's book on functional analysis. There you can find a proof of the open mapping theorem - that is Theorem 2.11. In fact, the idea of the proof is the same as in the case of banach spaces.
    – p4sch
    Dec 2 at 11:03


















  • why does "no finer topology" follow from the open mapping theorem?
    – yoshi
    Dec 1 at 15:22








  • 1




    If $tau_1$, $tau_2$ are topologies on $C^infty(mathbb{R})$ with $tau_1 subset tau_2$, i.e. $tau_2$ is finer than $tau_1$, then $mathrm{id} colon (C^infty(mathbb{R}),tau_2) rightarrow (C^infty(mathbb{R}),tau_1)$ the map is continuous. If both topologies are Frechet, then the open mapping theorem already implies that this map is open, i.e. the inverse map $mathrm{id} colon (C^infty(mathbb{R}),tau_1) rightarrow (C^infty(mathbb{R}),tau_2)$ is continuous and thus $tau_1 = tau_2$.
    – p4sch
    Dec 1 at 21:27












  • I like this, but my only problem is that we haven't covered the open mapping theorem for Frechet spaces. Is there a way to adapt Tito Aliatron's hints into this? I'm looking through the proof of the open mapping theorem for Frechet spaces to see if I can do it too.
    – yoshi
    Dec 1 at 21:44








  • 1




    I don't think that you can prove this much easier. It is a feature coming from the completeness of the space. You may have a look in Rudin's book on functional analysis. There you can find a proof of the open mapping theorem - that is Theorem 2.11. In fact, the idea of the proof is the same as in the case of banach spaces.
    – p4sch
    Dec 2 at 11:03
















why does "no finer topology" follow from the open mapping theorem?
– yoshi
Dec 1 at 15:22






why does "no finer topology" follow from the open mapping theorem?
– yoshi
Dec 1 at 15:22






1




1




If $tau_1$, $tau_2$ are topologies on $C^infty(mathbb{R})$ with $tau_1 subset tau_2$, i.e. $tau_2$ is finer than $tau_1$, then $mathrm{id} colon (C^infty(mathbb{R}),tau_2) rightarrow (C^infty(mathbb{R}),tau_1)$ the map is continuous. If both topologies are Frechet, then the open mapping theorem already implies that this map is open, i.e. the inverse map $mathrm{id} colon (C^infty(mathbb{R}),tau_1) rightarrow (C^infty(mathbb{R}),tau_2)$ is continuous and thus $tau_1 = tau_2$.
– p4sch
Dec 1 at 21:27






If $tau_1$, $tau_2$ are topologies on $C^infty(mathbb{R})$ with $tau_1 subset tau_2$, i.e. $tau_2$ is finer than $tau_1$, then $mathrm{id} colon (C^infty(mathbb{R}),tau_2) rightarrow (C^infty(mathbb{R}),tau_1)$ the map is continuous. If both topologies are Frechet, then the open mapping theorem already implies that this map is open, i.e. the inverse map $mathrm{id} colon (C^infty(mathbb{R}),tau_1) rightarrow (C^infty(mathbb{R}),tau_2)$ is continuous and thus $tau_1 = tau_2$.
– p4sch
Dec 1 at 21:27














I like this, but my only problem is that we haven't covered the open mapping theorem for Frechet spaces. Is there a way to adapt Tito Aliatron's hints into this? I'm looking through the proof of the open mapping theorem for Frechet spaces to see if I can do it too.
– yoshi
Dec 1 at 21:44






I like this, but my only problem is that we haven't covered the open mapping theorem for Frechet spaces. Is there a way to adapt Tito Aliatron's hints into this? I'm looking through the proof of the open mapping theorem for Frechet spaces to see if I can do it too.
– yoshi
Dec 1 at 21:44






1




1




I don't think that you can prove this much easier. It is a feature coming from the completeness of the space. You may have a look in Rudin's book on functional analysis. There you can find a proof of the open mapping theorem - that is Theorem 2.11. In fact, the idea of the proof is the same as in the case of banach spaces.
– p4sch
Dec 2 at 11:03




I don't think that you can prove this much easier. It is a feature coming from the completeness of the space. You may have a look in Rudin's book on functional analysis. There you can find a proof of the open mapping theorem - that is Theorem 2.11. In fact, the idea of the proof is the same as in the case of banach spaces.
– p4sch
Dec 2 at 11:03


















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Berounka

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Berounka.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Caractéristiques Longueur 139  km Bassin 8 861  km 2 Bassin collecteur Elbe Débit moyen ? Cours Source Confluence de la Mže et de la Radbuza · Localisation Pilsen · Altitude 305  m · Coordonnées 49° 45′ 13″ N, 13° 23′ 24″ E Confluence Vltava · Localisation Prague · Altitude 187  m · Coordonnées 49° 59′ 43″ N, 14° 24′ 04″ E Géographie Pays traversés   République tchèque modifier   La Berounka est une rivière de la République tchèque. Elle porte le nom de Mže depuis sa source en Allemagne près de la frontière jusqu'à sa confluence avec la Radbuza à Plzeň. Elle continue ensuite sous le nom de Berounka jusqu'à sa confluence avec la Vltava, près de Prague. La rivière est une destination prisée pour les amateurs de canoë, qui appréc...
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Sphinx de Gizeh

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Pour les articles homonymes, voir Sphinx. Sphinx de Gizeh.mw-parser-output .entete.egypte-antique{background-image:url("//upload.wikimedia.org/wikipedia/commons/c/cc/Picto_infobox_ancient_Egypt.png")} Site d'Égypte antique Le visage du sphinx de Gizeh. Noms en arabe أبو الهول , Abou al-Hôl Localisation Région Basse-Égypte Coordonnées 29° 58′ 31″ nord, 31° 08′ 16″ est Géolocalisation sur la carte : Égypte Sphinx de Gizeh modifier   Le sphinx de Gizeh est la statue thérianthrope qui se dresse devant les grandes pyramides du plateau de Gizeh, en Basse-Égypte. Sculpture monumentale monolithique la plus grande du monde [ 1 ] avec 73,5 mètres de longueur, 14 mètres de largeur et 20,22 mètres de hauteur, elle représente un sphinx couchant. Réalisée vers -2500, elle serait attribuée à Djédefrê, l'un des pharaons de la IV e  dynastie, représentant son père le pharaon Khéops. Sommaire 1 Étymologie 2 Desc...
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Different font size/position of beamer's navigation symbols template's content depending on regular/plain...

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3 Let's suppose I'd like to divert the navigation symbols template (e.g. in order replace them by the frame numbers, see this bug report ;) I would be facing troubles regarding the font size/position, depending on regular/plain frames, as shown by the following MCE. How could I get rid of these troubles? documentclass{beamer} usetheme{PaloAlto} setbeamertemplate{navigation symbols}{insertframenumber/inserttotalframenumber} begin{document} begin{frame} frametitle{A regular frame} end{frame} begin{frame}[plain] frametitle{A plain frame} end{frame} end{document} beamer header-footer share | improve this question asked Dec 10 '18 a...
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