Frechet derivative of a function on matrix-valued $L^2$ functions
$begingroup$
$newcommand{tr}{text{Tr}}$
Let us denote $L^2(X, mathbb C^{n times n})$ be the space of matrix-valued $L^2$ functions. That is, if $f in L^2(X, mathbb C^{n times n})$, each entry of $f$ is an $L^2$ function, i.e., $f_{ij} in L^2(X, mathbb C)$ for $i,j in {1, dots, n}$. Suppose we have a fixed $g in L^2(X, mathbb C^{n times n})$, we define a function $F: L^2(X, mathbb C^{n times n}) to mathbb R$ by
$$ f mapsto text{Tr}(f^*g) mapsto left( int|text{Tr}(f^*g))|^2 dx right).$$
Is $F$ Frechet differentiable? I did following computation
begin{align*}
F(f+h) - F(f) &= int tr(g^*(f+h)) tr((f+h)^*g) - int tr(g^*f)tr(f^*g) \
&=int tr(g^* h)tr(f^*g) + inttr(g^*h)tr(h^*g) + int tr(g^*f)tr(h^*g).
end{align*}
Now letting $h to 0$ in $L^2$, I am not sure the quotient
begin{align*}
lim_{|h|_{L^2} to 0} frac{|F(f+h) - F(f)|}{|h|_{L^2}}
end{align*}
converges to something.
real-analysis lp-spaces frechet-derivative
asked Dec 12 '18 at 0:13
user1101010user1101010
7191730
$endgroup$
|
show 2 more comments
$begingroup$
$newcommand{tr}{text{Tr}}$
Let us denote $L^2(X, mathbb C^{n times n})$ be the space of matrix-valued $L^2$ functions. That is, if $f in L^2(X, mathbb C^{n times n})$, each entry of $f$ is an $L^2$ function, i.e., $f_{ij} in L^2(X, mathbb C)$ for $i,j in {1, dots, n}$. Suppose we have a fixed $g in L^2(X, mathbb C^{n times n})$, we define a function $F: L^2(X, mathbb C^{n times n}) to mathbb R$ by
$$ f mapsto text{Tr}(f^*g) mapsto left( int|text{Tr}(f^*g))|^2 dx right).$$
Is $F$ Frechet differentiable? I did following computation
begin{align*}
F(f+h) - F(f) &= int tr(g^*(f+h)) tr((f+h)^*g) - int tr(g^*f)tr(f^*g) \
&=int tr(g^* h)tr(f^*g) + inttr(g^*h)tr(h^*g) + int tr(g^*f)tr(h^*g).
end{align*}
Now letting $h to 0$ in $L^2$, I am not sure the quotient
begin{align*}
lim_{|h|_{L^2} to 0} frac{|F(f+h) - F(f)|}{|h|_{L^2}}
end{align*}
converges to something.
real-analysis lp-spaces frechet-derivative
asked Dec 12 '18 at 0:13
user1101010user1101010
7191730
$endgroup$
$begingroup$
Try scalar valued functions first with $g=1$.
$endgroup$
– copper.hat
Dec 12 '18 at 4:23
$begingroup$
@copper.hat: With $g=1$, is it that $F$ just a linear functional over $L^2(X)$? So in this case, does it mean $DF(f)[h] = int h dx$?
$endgroup$
– user1101010
Dec 12 '18 at 5:04
$begingroup$
There is a square of modulus up there...
$endgroup$
– copper.hat
Dec 12 '18 at 5:11
$begingroup$
@copper.hat: I see, thanks. In that case, $F(f) = |f|_{L^2}^2$. Then $F(f+h) - F(h) = langle f, h rangle + langle h, f rangle + langle h, h rangle$. I think $frac{ langle h, h rangle }{|h|_{L^2} }to 0 $ as $|h|_{L^2} to 0$. Then is it true $DF(f)[h] = int f^* h dx + int h^* f dx$. This should define a bounded linear functional on $L^2$ and so it is Frechet differentiable?
$endgroup$
– user1101010
Dec 12 '18 at 5:19
$begingroup$
Unfortunately it is not differentiable because of the complex part. If it was a real valued functional it would be. For the same reason that $z mapsto |z|^2$ is not analytic.
$endgroup$
– copper.hat
Dec 12 '18 at 5:50
|
show 2 more comments
$begingroup$
$newcommand{tr}{text{Tr}}$
Let us denote $L^2(X, mathbb C^{n times n})$ be the space of matrix-valued $L^2$ functions. That is, if $f in L^2(X, mathbb C^{n times n})$, each entry of $f$ is an $L^2$ function, i.e., $f_{ij} in L^2(X, mathbb C)$ for $i,j in {1, dots, n}$. Suppose we have a fixed $g in L^2(X, mathbb C^{n times n})$, we define a function $F: L^2(X, mathbb C^{n times n}) to mathbb R$ by
$$ f mapsto text{Tr}(f^*g) mapsto left( int|text{Tr}(f^*g))|^2 dx right).$$
Is $F$ Frechet differentiable? I did following computation
begin{align*}
F(f+h) - F(f) &= int tr(g^*(f+h)) tr((f+h)^*g) - int tr(g^*f)tr(f^*g) \
&=int tr(g^* h)tr(f^*g) + inttr(g^*h)tr(h^*g) + int tr(g^*f)tr(h^*g).
end{align*}
Now letting $h to 0$ in $L^2$, I am not sure the quotient
begin{align*}
lim_{|h|_{L^2} to 0} frac{|F(f+h) - F(f)|}{|h|_{L^2}}
end{align*}
converges to something.
real-analysis lp-spaces frechet-derivative
asked Dec 12 '18 at 0:13
user1101010user1101010
7191730
$endgroup$
$newcommand{tr}{text{Tr}}$
Let us denote $L^2(X, mathbb C^{n times n})$ be the space of matrix-valued $L^2$ functions. That is, if $f in L^2(X, mathbb C^{n times n})$, each entry of $f$ is an $L^2$ function, i.e., $f_{ij} in L^2(X, mathbb C)$ for $i,j in {1, dots, n}$. Suppose we have a fixed $g in L^2(X, mathbb C^{n times n})$, we define a function $F: L^2(X, mathbb C^{n times n}) to mathbb R$ by
$$ f mapsto text{Tr}(f^*g) mapsto left( int|text{Tr}(f^*g))|^2 dx right).$$
Is $F$ Frechet differentiable? I did following computation
begin{align*}
F(f+h) - F(f) &= int tr(g^*(f+h)) tr((f+h)^*g) - int tr(g^*f)tr(f^*g) \
&=int tr(g^* h)tr(f^*g) + inttr(g^*h)tr(h^*g) + int tr(g^*f)tr(h^*g).
end{align*}
Now letting $h to 0$ in $L^2$, I am not sure the quotient
begin{align*}
lim_{|h|_{L^2} to 0} frac{|F(f+h) - F(f)|}{|h|_{L^2}}
end{align*}
converges to something.
real-analysis lp-spaces frechet-derivative
real-analysis lp-spaces frechet-derivative
asked Dec 12 '18 at 0:13
user1101010user1101010
7191730
asked Dec 12 '18 at 0:13
user1101010user1101010
7191730
asked Dec 12 '18 at 0:13
user1101010user1101010
7191730
asked Dec 12 '18 at 0:13
user1101010user1101010
7191730
asked Dec 12 '18 at 0:13
user1101010user1101010
7191730
7191730
$begingroup$
Try scalar valued functions first with $g=1$.
$endgroup$
– copper.hat
Dec 12 '18 at 4:23
$begingroup$
@copper.hat: With $g=1$, is it that $F$ just a linear functional over $L^2(X)$? So in this case, does it mean $DF(f)[h] = int h dx$?
$endgroup$
– user1101010
Dec 12 '18 at 5:04
$begingroup$
There is a square of modulus up there...
$endgroup$
– copper.hat
Dec 12 '18 at 5:11
$begingroup$
@copper.hat: I see, thanks. In that case, $F(f) = |f|_{L^2}^2$. Then $F(f+h) - F(h) = langle f, h rangle + langle h, f rangle + langle h, h rangle$. I think $frac{ langle h, h rangle }{|h|_{L^2} }to 0 $ as $|h|_{L^2} to 0$. Then is it true $DF(f)[h] = int f^* h dx + int h^* f dx$. This should define a bounded linear functional on $L^2$ and so it is Frechet differentiable?
$endgroup$
– user1101010
Dec 12 '18 at 5:19
$begingroup$
Unfortunately it is not differentiable because of the complex part. If it was a real valued functional it would be. For the same reason that $z mapsto |z|^2$ is not analytic.
$endgroup$
– copper.hat
Dec 12 '18 at 5:50
|
show 2 more comments
$begingroup$
Try scalar valued functions first with $g=1$.
$endgroup$
– copper.hat
Dec 12 '18 at 4:23
$begingroup$
@copper.hat: With $g=1$, is it that $F$ just a linear functional over $L^2(X)$? So in this case, does it mean $DF(f)[h] = int h dx$?
$endgroup$
– user1101010
Dec 12 '18 at 5:04
$begingroup$
There is a square of modulus up there...
$endgroup$
– copper.hat
Dec 12 '18 at 5:11
$begingroup$
@copper.hat: I see, thanks. In that case, $F(f) = |f|_{L^2}^2$. Then $F(f+h) - F(h) = langle f, h rangle + langle h, f rangle + langle h, h rangle$. I think $frac{ langle h, h rangle }{|h|_{L^2} }to 0 $ as $|h|_{L^2} to 0$. Then is it true $DF(f)[h] = int f^* h dx + int h^* f dx$. This should define a bounded linear functional on $L^2$ and so it is Frechet differentiable?
$endgroup$
– user1101010
Dec 12 '18 at 5:19
$begingroup$
Unfortunately it is not differentiable because of the complex part. If it was a real valued functional it would be. For the same reason that $z mapsto |z|^2$ is not analytic.
$endgroup$
– copper.hat
Dec 12 '18 at 5:50
$begingroup$
Try scalar valued functions first with $g=1$.
$endgroup$
– copper.hat
Dec 12 '18 at 4:23
$begingroup$
Try scalar valued functions first with $g=1$.
$endgroup$
– copper.hat
Dec 12 '18 at 4:23
$begingroup$
@copper.hat: With $g=1$, is it that $F$ just a linear functional over $L^2(X)$? So in this case, does it mean $DF(f)[h] = int h dx$?
$endgroup$
– user1101010
Dec 12 '18 at 5:04
$begingroup$
@copper.hat: With $g=1$, is it that $F$ just a linear functional over $L^2(X)$? So in this case, does it mean $DF(f)[h] = int h dx$?
$endgroup$
– user1101010
Dec 12 '18 at 5:04
$begingroup$
There is a square of modulus up there...
$endgroup$
– copper.hat
Dec 12 '18 at 5:11
$begingroup$
There is a square of modulus up there...
$endgroup$
– copper.hat
Dec 12 '18 at 5:11
$begingroup$
@copper.hat: I see, thanks. In that case, $F(f) = |f|_{L^2}^2$. Then $F(f+h) - F(h) = langle f, h rangle + langle h, f rangle + langle h, h rangle$. I think $frac{ langle h, h rangle }{|h|_{L^2} }to 0 $ as $|h|_{L^2} to 0$. Then is it true $DF(f)[h] = int f^* h dx + int h^* f dx$. This should define a bounded linear functional on $L^2$ and so it is Frechet differentiable?
$endgroup$
– user1101010
Dec 12 '18 at 5:19
$begingroup$
@copper.hat: I see, thanks. In that case, $F(f) = |f|_{L^2}^2$. Then $F(f+h) - F(h) = langle f, h rangle + langle h, f rangle + langle h, h rangle$. I think $frac{ langle h, h rangle }{|h|_{L^2} }to 0 $ as $|h|_{L^2} to 0$. Then is it true $DF(f)[h] = int f^* h dx + int h^* f dx$. This should define a bounded linear functional on $L^2$ and so it is Frechet differentiable?
$endgroup$
– user1101010
Dec 12 '18 at 5:19
$begingroup$
Unfortunately it is not differentiable because of the complex part. If it was a real valued functional it would be. For the same reason that $z mapsto |z|^2$ is not analytic.
$endgroup$
– copper.hat
Dec 12 '18 at 5:50
$begingroup$
Unfortunately it is not differentiable because of the complex part. If it was a real valued functional it would be. For the same reason that $z mapsto |z|^2$ is not analytic.
$endgroup$
– copper.hat
Dec 12 '18 at 5:50
|
show 2 more comments
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$begingroup$
Try scalar valued functions first with $g=1$.
$endgroup$
– copper.hat
Dec 12 '18 at 4:23
$begingroup$
@copper.hat: With $g=1$, is it that $F$ just a linear functional over $L^2(X)$? So in this case, does it mean $DF(f)[h] = int h dx$?
$endgroup$
– user1101010
Dec 12 '18 at 5:04
$begingroup$
There is a square of modulus up there...
$endgroup$
– copper.hat
Dec 12 '18 at 5:11
$begingroup$
@copper.hat: I see, thanks. In that case, $F(f) = |f|_{L^2}^2$. Then $F(f+h) - F(h) = langle f, h rangle + langle h, f rangle + langle h, h rangle$. I think $frac{ langle h, h rangle }{|h|_{L^2} }to 0 $ as $|h|_{L^2} to 0$. Then is it true $DF(f)[h] = int f^* h dx + int h^* f dx$. This should define a bounded linear functional on $L^2$ and so it is Frechet differentiable?
$endgroup$
– user1101010
Dec 12 '18 at 5:19
$begingroup$
Unfortunately it is not differentiable because of the complex part. If it was a real valued functional it would be. For the same reason that $z mapsto |z|^2$ is not analytic.
$endgroup$
– copper.hat
Dec 12 '18 at 5:50