For a bounded linear functional $l(v) := (f,v)_Omega$, do we have $|l| le |f|$?
Suppose we have a Poisson equation $-Delta u = f$ on $Omega$ and we want to derive its weak formulation, so
we multiply it by an arbitrary test function $forall v in H^1_0$ and then take integral over $Omega$
begin{aligned}
(-Delta u,v)_Omega = (f,v)_Omega
end{aligned}
where we denote by $(cdot,cdot)_Omega$ the inner product, and we want to find the weak solution $u in H^1_0$ such that
begin{aligned}
a(u,v) = l(v)
end{aligned}
where the bilinear form $a(u,v) := (-Delta u,v)_Omega$ and the linear form $l(v) := (f,v)_Omega$.
Let's focus on the right-hand side. By Cauchy-Schwarz inequality we have $(f,v)_Omega le |f||v|$, while by the boundedness of the linear functional we have $l(v) le |l||v|$, where$|l|:= sup_{|v|=1} |l(v)|$ denotes the operator norm. Note that we also have $(f,v)_Omega=l(v)$. We may thus conclude that $|l| le |f|$. Is that correct? Can we further obtain $|l|=|f|$?
functional-analysis pde numerical-methods bilinear-form finite-element-method
asked Nov 26 at 14:49
Analysis Newbie
41627
add a comment |
Suppose we have a Poisson equation $-Delta u = f$ on $Omega$ and we want to derive its weak formulation, so
we multiply it by an arbitrary test function $forall v in H^1_0$ and then take integral over $Omega$
begin{aligned}
(-Delta u,v)_Omega = (f,v)_Omega
end{aligned}
where we denote by $(cdot,cdot)_Omega$ the inner product, and we want to find the weak solution $u in H^1_0$ such that
begin{aligned}
a(u,v) = l(v)
end{aligned}
where the bilinear form $a(u,v) := (-Delta u,v)_Omega$ and the linear form $l(v) := (f,v)_Omega$.
Let's focus on the right-hand side. By Cauchy-Schwarz inequality we have $(f,v)_Omega le |f||v|$, while by the boundedness of the linear functional we have $l(v) le |l||v|$, where$|l|:= sup_{|v|=1} |l(v)|$ denotes the operator norm. Note that we also have $(f,v)_Omega=l(v)$. We may thus conclude that $|l| le |f|$. Is that correct? Can we further obtain $|l|=|f|$?
functional-analysis pde numerical-methods bilinear-form finite-element-method
asked Nov 26 at 14:49
Analysis Newbie
41627
$|ell|leq |f|$ is indeed obvious (and you justified it quite correctly). But be careful nevertheless : $aleq |b|$ doesn't implies that $|a|leq |b|$. Be here $ell$ is linear so you can manage the lower bound). For the converse inequality (that is also true), If $E$ is a banach space, then one can prove that $|x|_E=sup_{|f|_{E'}leq 1}|f(x)|,$ where $E'$ is the topological dual of $E$.
– Surb
Nov 26 at 14:58
By definition, $l(v) := (f,v)_Omega$. The estimate $|l(v)|=|(f,v)_Omega| le |f||v|$ proves two things: First, $l$ is bounded. Second, $|l|leq |f|$.
– Pedro
Nov 26 at 15:51
add a comment |
Suppose we have a Poisson equation $-Delta u = f$ on $Omega$ and we want to derive its weak formulation, so
we multiply it by an arbitrary test function $forall v in H^1_0$ and then take integral over $Omega$
begin{aligned}
(-Delta u,v)_Omega = (f,v)_Omega
end{aligned}
where we denote by $(cdot,cdot)_Omega$ the inner product, and we want to find the weak solution $u in H^1_0$ such that
begin{aligned}
a(u,v) = l(v)
end{aligned}
where the bilinear form $a(u,v) := (-Delta u,v)_Omega$ and the linear form $l(v) := (f,v)_Omega$.
Let's focus on the right-hand side. By Cauchy-Schwarz inequality we have $(f,v)_Omega le |f||v|$, while by the boundedness of the linear functional we have $l(v) le |l||v|$, where$|l|:= sup_{|v|=1} |l(v)|$ denotes the operator norm. Note that we also have $(f,v)_Omega=l(v)$. We may thus conclude that $|l| le |f|$. Is that correct? Can we further obtain $|l|=|f|$?
functional-analysis pde numerical-methods bilinear-form finite-element-method
asked Nov 26 at 14:49
Analysis Newbie
41627
Suppose we have a Poisson equation $-Delta u = f$ on $Omega$ and we want to derive its weak formulation, so
we multiply it by an arbitrary test function $forall v in H^1_0$ and then take integral over $Omega$
begin{aligned}
(-Delta u,v)_Omega = (f,v)_Omega
end{aligned}
where we denote by $(cdot,cdot)_Omega$ the inner product, and we want to find the weak solution $u in H^1_0$ such that
begin{aligned}
a(u,v) = l(v)
end{aligned}
where the bilinear form $a(u,v) := (-Delta u,v)_Omega$ and the linear form $l(v) := (f,v)_Omega$.
Let's focus on the right-hand side. By Cauchy-Schwarz inequality we have $(f,v)_Omega le |f||v|$, while by the boundedness of the linear functional we have $l(v) le |l||v|$, where$|l|:= sup_{|v|=1} |l(v)|$ denotes the operator norm. Note that we also have $(f,v)_Omega=l(v)$. We may thus conclude that $|l| le |f|$. Is that correct? Can we further obtain $|l|=|f|$?
functional-analysis pde numerical-methods bilinear-form finite-element-method
functional-analysis pde numerical-methods bilinear-form finite-element-method
asked Nov 26 at 14:49
Analysis Newbie
41627
asked Nov 26 at 14:49
Analysis Newbie
41627
asked Nov 26 at 14:49
Analysis Newbie
41627
asked Nov 26 at 14:49
Analysis Newbie
41627
asked Nov 26 at 14:49
Analysis Newbie
41627
41627
$|ell|leq |f|$ is indeed obvious (and you justified it quite correctly). But be careful nevertheless : $aleq |b|$ doesn't implies that $|a|leq |b|$. Be here $ell$ is linear so you can manage the lower bound). For the converse inequality (that is also true), If $E$ is a banach space, then one can prove that $|x|_E=sup_{|f|_{E'}leq 1}|f(x)|,$ where $E'$ is the topological dual of $E$.
– Surb
Nov 26 at 14:58
By definition, $l(v) := (f,v)_Omega$. The estimate $|l(v)|=|(f,v)_Omega| le |f||v|$ proves two things: First, $l$ is bounded. Second, $|l|leq |f|$.
– Pedro
Nov 26 at 15:51
add a comment |
$|ell|leq |f|$ is indeed obvious (and you justified it quite correctly). But be careful nevertheless : $aleq |b|$ doesn't implies that $|a|leq |b|$. Be here $ell$ is linear so you can manage the lower bound). For the converse inequality (that is also true), If $E$ is a banach space, then one can prove that $|x|_E=sup_{|f|_{E'}leq 1}|f(x)|,$ where $E'$ is the topological dual of $E$.
– Surb
Nov 26 at 14:58
By definition, $l(v) := (f,v)_Omega$. The estimate $|l(v)|=|(f,v)_Omega| le |f||v|$ proves two things: First, $l$ is bounded. Second, $|l|leq |f|$.
– Pedro
Nov 26 at 15:51
$|ell|leq |f|$ is indeed obvious (and you justified it quite correctly). But be careful nevertheless : $aleq |b|$ doesn't implies that $|a|leq |b|$. Be here $ell$ is linear so you can manage the lower bound). For the converse inequality (that is also true), If $E$ is a banach space, then one can prove that $|x|_E=sup_{|f|_{E'}leq 1}|f(x)|,$ where $E'$ is the topological dual of $E$.
– Surb
Nov 26 at 14:58
$|ell|leq |f|$ is indeed obvious (and you justified it quite correctly). But be careful nevertheless : $aleq |b|$ doesn't implies that $|a|leq |b|$. Be here $ell$ is linear so you can manage the lower bound). For the converse inequality (that is also true), If $E$ is a banach space, then one can prove that $|x|_E=sup_{|f|_{E'}leq 1}|f(x)|,$ where $E'$ is the topological dual of $E$.
– Surb
Nov 26 at 14:58
By definition, $l(v) := (f,v)_Omega$. The estimate $|l(v)|=|(f,v)_Omega| le |f||v|$ proves two things: First, $l$ is bounded. Second, $|l|leq |f|$.
– Pedro
Nov 26 at 15:51
By definition, $l(v) := (f,v)_Omega$. The estimate $|l(v)|=|(f,v)_Omega| le |f||v|$ proves two things: First, $l$ is bounded. Second, $|l|leq |f|$.
– Pedro
Nov 26 at 15:51
add a comment |
1 Answer
1
active
oldest
votes
Yes, this is correct.
However, I would recommend to specify the norms for $f$ and $v$ in your proof (I assume you mean the $L^2(Omega)$ norm). For $l$ it is clear in my opinion that the norm refers to the operator norm.
Further, we can obtain $|l|=|f|_{L^2(Omega)}$.
This can be seen by choosing $v:=fin L^2(Omega)$.
Then we have
$$
| l ||f|_{L^2(Omega)} geq |l(f)| = (f,f)_Omega = |f|^2_{L^2(Omega)}.
$$
Dividing by $|f|_{L^2(Omega)}$ yields the result (you have to consider the special case $f=0$, which is trivial).
answered Nov 26 at 15:00
supinf
5,9591027
Thank you very much! Is there any theorem related to this result (i.e., $|l|=|f|$) ? I am thinking about the Riesz representation theorem that says, we can always find a unique $f in H$ such that $l_f(v) = (f,v)$ for $forall v in H$ and $l in H^*$, and $|l|=|f|$. So can we say our result is a consequence of the Riesz representation theorem?
– Analysis Newbie
Nov 26 at 17:53
yes, I would say this is the case.
– supinf
Nov 27 at 10:13
add a comment |
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Yes, this is correct.
However, I would recommend to specify the norms for $f$ and $v$ in your proof (I assume you mean the $L^2(Omega)$ norm). For $l$ it is clear in my opinion that the norm refers to the operator norm.
Further, we can obtain $|l|=|f|_{L^2(Omega)}$.
This can be seen by choosing $v:=fin L^2(Omega)$.
Then we have
$$
| l ||f|_{L^2(Omega)} geq |l(f)| = (f,f)_Omega = |f|^2_{L^2(Omega)}.
$$
Dividing by $|f|_{L^2(Omega)}$ yields the result (you have to consider the special case $f=0$, which is trivial).
answered Nov 26 at 15:00
supinf
5,9591027
Thank you very much! Is there any theorem related to this result (i.e., $|l|=|f|$) ? I am thinking about the Riesz representation theorem that says, we can always find a unique $f in H$ such that $l_f(v) = (f,v)$ for $forall v in H$ and $l in H^*$, and $|l|=|f|$. So can we say our result is a consequence of the Riesz representation theorem?
– Analysis Newbie
Nov 26 at 17:53
yes, I would say this is the case.
– supinf
Nov 27 at 10:13
add a comment |
Yes, this is correct.
However, I would recommend to specify the norms for $f$ and $v$ in your proof (I assume you mean the $L^2(Omega)$ norm). For $l$ it is clear in my opinion that the norm refers to the operator norm.
Further, we can obtain $|l|=|f|_{L^2(Omega)}$.
This can be seen by choosing $v:=fin L^2(Omega)$.
Then we have
$$
| l ||f|_{L^2(Omega)} geq |l(f)| = (f,f)_Omega = |f|^2_{L^2(Omega)}.
$$
Dividing by $|f|_{L^2(Omega)}$ yields the result (you have to consider the special case $f=0$, which is trivial).
answered Nov 26 at 15:00
supinf
5,9591027
Thank you very much! Is there any theorem related to this result (i.e., $|l|=|f|$) ? I am thinking about the Riesz representation theorem that says, we can always find a unique $f in H$ such that $l_f(v) = (f,v)$ for $forall v in H$ and $l in H^*$, and $|l|=|f|$. So can we say our result is a consequence of the Riesz representation theorem?
– Analysis Newbie
Nov 26 at 17:53
yes, I would say this is the case.
– supinf
Nov 27 at 10:13
add a comment |
Yes, this is correct.
However, I would recommend to specify the norms for $f$ and $v$ in your proof (I assume you mean the $L^2(Omega)$ norm). For $l$ it is clear in my opinion that the norm refers to the operator norm.
Further, we can obtain $|l|=|f|_{L^2(Omega)}$.
This can be seen by choosing $v:=fin L^2(Omega)$.
Then we have
$$
| l ||f|_{L^2(Omega)} geq |l(f)| = (f,f)_Omega = |f|^2_{L^2(Omega)}.
$$
Dividing by $|f|_{L^2(Omega)}$ yields the result (you have to consider the special case $f=0$, which is trivial).
answered Nov 26 at 15:00
supinf
5,9591027
Yes, this is correct.
However, I would recommend to specify the norms for $f$ and $v$ in your proof (I assume you mean the $L^2(Omega)$ norm). For $l$ it is clear in my opinion that the norm refers to the operator norm.
Further, we can obtain $|l|=|f|_{L^2(Omega)}$.
This can be seen by choosing $v:=fin L^2(Omega)$.
Then we have
$$
| l ||f|_{L^2(Omega)} geq |l(f)| = (f,f)_Omega = |f|^2_{L^2(Omega)}.
$$
Dividing by $|f|_{L^2(Omega)}$ yields the result (you have to consider the special case $f=0$, which is trivial).
answered Nov 26 at 15:00
supinf
5,9591027
answered Nov 26 at 15:00
supinf
5,9591027
answered Nov 26 at 15:00
supinf
5,9591027
answered Nov 26 at 15:00
supinf
5,9591027
5,9591027
Thank you very much! Is there any theorem related to this result (i.e., $|l|=|f|$) ? I am thinking about the Riesz representation theorem that says, we can always find a unique $f in H$ such that $l_f(v) = (f,v)$ for $forall v in H$ and $l in H^*$, and $|l|=|f|$. So can we say our result is a consequence of the Riesz representation theorem?
– Analysis Newbie
Nov 26 at 17:53
yes, I would say this is the case.
– supinf
Nov 27 at 10:13
add a comment |
Thank you very much! Is there any theorem related to this result (i.e., $|l|=|f|$) ? I am thinking about the Riesz representation theorem that says, we can always find a unique $f in H$ such that $l_f(v) = (f,v)$ for $forall v in H$ and $l in H^*$, and $|l|=|f|$. So can we say our result is a consequence of the Riesz representation theorem?
– Analysis Newbie
Nov 26 at 17:53
yes, I would say this is the case.
– supinf
Nov 27 at 10:13
Thank you very much! Is there any theorem related to this result (i.e., $|l|=|f|$) ? I am thinking about the Riesz representation theorem that says, we can always find a unique $f in H$ such that $l_f(v) = (f,v)$ for $forall v in H$ and $l in H^*$, and $|l|=|f|$. So can we say our result is a consequence of the Riesz representation theorem?
– Analysis Newbie
Nov 26 at 17:53
Thank you very much! Is there any theorem related to this result (i.e., $|l|=|f|$) ? I am thinking about the Riesz representation theorem that says, we can always find a unique $f in H$ such that $l_f(v) = (f,v)$ for $forall v in H$ and $l in H^*$, and $|l|=|f|$. So can we say our result is a consequence of the Riesz representation theorem?
– Analysis Newbie
Nov 26 at 17:53
yes, I would say this is the case.
– supinf
Nov 27 at 10:13
yes, I would say this is the case.
– supinf
Nov 27 at 10:13
add a comment |
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$|ell|leq |f|$ is indeed obvious (and you justified it quite correctly). But be careful nevertheless : $aleq |b|$ doesn't implies that $|a|leq |b|$. Be here $ell$ is linear so you can manage the lower bound). For the converse inequality (that is also true), If $E$ is a banach space, then one can prove that $|x|_E=sup_{|f|_{E'}leq 1}|f(x)|,$ where $E'$ is the topological dual of $E$.
– Surb
Nov 26 at 14:58
By definition, $l(v) := (f,v)_Omega$. The estimate $|l(v)|=|(f,v)_Omega| le |f||v|$ proves two things: First, $l$ is bounded. Second, $|l|leq |f|$.
– Pedro
Nov 26 at 15:51