Approximate a continuous functions with $C^infty$ functions in uniform norm.
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I have a question concerning the generator of a backwards SDE. Suppose that $f(omega,t,y,z):Omegatimesmathbb{R}^+timesmathbb{R}timesmathbb{R}^drightarrowmathbb{R}$ is almost surely jointly continuous in $(t,y,z)$. Suppose further that $f$ has suitable growth conditions (e.g. linear in $y$ and quadratic in $z$.
In the Kobylanski 2000 paper, it was asserted that there is a sequence $f^{(n)}$ of $C^infty$ functions such that:
$$f+frac{1}{2^{n+1}}leq f^{(n)} leq f+frac{1}{2^n}$$
I'm at a bit of a loss as to how to prove the existence of such a sequence. In the said paper, it was said that the existence of this sequence is given by a standard argument of regularisation, but I don't quite see how. Any help or reference would be greatly appreciated. Thanks!
real-analysis probability functional-analysis probability-theory stochastic-calculus
asked Jan 25 at 11:06
Viet DangViet Dang
162
$endgroup$
add a comment |
$begingroup$
I have a question concerning the generator of a backwards SDE. Suppose that $f(omega,t,y,z):Omegatimesmathbb{R}^+timesmathbb{R}timesmathbb{R}^drightarrowmathbb{R}$ is almost surely jointly continuous in $(t,y,z)$. Suppose further that $f$ has suitable growth conditions (e.g. linear in $y$ and quadratic in $z$.
In the Kobylanski 2000 paper, it was asserted that there is a sequence $f^{(n)}$ of $C^infty$ functions such that:
$$f+frac{1}{2^{n+1}}leq f^{(n)} leq f+frac{1}{2^n}$$
I'm at a bit of a loss as to how to prove the existence of such a sequence. In the said paper, it was said that the existence of this sequence is given by a standard argument of regularisation, but I don't quite see how. Any help or reference would be greatly appreciated. Thanks!
real-analysis probability functional-analysis probability-theory stochastic-calculus
asked Jan 25 at 11:06
Viet DangViet Dang
162
$endgroup$
$begingroup$
If $f$ is uniformly continuous, you can take $deltain (2^{-(n+1)},2^{-n})$ and convolute $f+delta$ with a mollifier $eta_epsilon$. The resulting function is smooth and converges uniformly to $f+delta$ as $epsilonsearrow 0$. One can probably make something similar work with the assumption of uniform continuity.
$endgroup$
– MaoWao
Jan 25 at 11:30
add a comment |
$begingroup$
I have a question concerning the generator of a backwards SDE. Suppose that $f(omega,t,y,z):Omegatimesmathbb{R}^+timesmathbb{R}timesmathbb{R}^drightarrowmathbb{R}$ is almost surely jointly continuous in $(t,y,z)$. Suppose further that $f$ has suitable growth conditions (e.g. linear in $y$ and quadratic in $z$.
In the Kobylanski 2000 paper, it was asserted that there is a sequence $f^{(n)}$ of $C^infty$ functions such that:
$$f+frac{1}{2^{n+1}}leq f^{(n)} leq f+frac{1}{2^n}$$
I'm at a bit of a loss as to how to prove the existence of such a sequence. In the said paper, it was said that the existence of this sequence is given by a standard argument of regularisation, but I don't quite see how. Any help or reference would be greatly appreciated. Thanks!
real-analysis probability functional-analysis probability-theory stochastic-calculus
asked Jan 25 at 11:06
Viet DangViet Dang
162
$endgroup$
I have a question concerning the generator of a backwards SDE. Suppose that $f(omega,t,y,z):Omegatimesmathbb{R}^+timesmathbb{R}timesmathbb{R}^drightarrowmathbb{R}$ is almost surely jointly continuous in $(t,y,z)$. Suppose further that $f$ has suitable growth conditions (e.g. linear in $y$ and quadratic in $z$.
In the Kobylanski 2000 paper, it was asserted that there is a sequence $f^{(n)}$ of $C^infty$ functions such that:
$$f+frac{1}{2^{n+1}}leq f^{(n)} leq f+frac{1}{2^n}$$
I'm at a bit of a loss as to how to prove the existence of such a sequence. In the said paper, it was said that the existence of this sequence is given by a standard argument of regularisation, but I don't quite see how. Any help or reference would be greatly appreciated. Thanks!
real-analysis probability functional-analysis probability-theory stochastic-calculus
real-analysis probability functional-analysis probability-theory stochastic-calculus
asked Jan 25 at 11:06
Viet DangViet Dang
162
asked Jan 25 at 11:06
Viet DangViet Dang
162
asked Jan 25 at 11:06
Viet DangViet Dang
162
asked Jan 25 at 11:06
Viet DangViet Dang
162
asked Jan 25 at 11:06
Viet DangViet Dang
162
162
$begingroup$
If $f$ is uniformly continuous, you can take $deltain (2^{-(n+1)},2^{-n})$ and convolute $f+delta$ with a mollifier $eta_epsilon$. The resulting function is smooth and converges uniformly to $f+delta$ as $epsilonsearrow 0$. One can probably make something similar work with the assumption of uniform continuity.
$endgroup$
– MaoWao
Jan 25 at 11:30
add a comment |
$begingroup$
If $f$ is uniformly continuous, you can take $deltain (2^{-(n+1)},2^{-n})$ and convolute $f+delta$ with a mollifier $eta_epsilon$. The resulting function is smooth and converges uniformly to $f+delta$ as $epsilonsearrow 0$. One can probably make something similar work with the assumption of uniform continuity.
$endgroup$
– MaoWao
Jan 25 at 11:30
$begingroup$
If $f$ is uniformly continuous, you can take $deltain (2^{-(n+1)},2^{-n})$ and convolute $f+delta$ with a mollifier $eta_epsilon$. The resulting function is smooth and converges uniformly to $f+delta$ as $epsilonsearrow 0$. One can probably make something similar work with the assumption of uniform continuity.
$endgroup$
– MaoWao
Jan 25 at 11:30
$begingroup$
If $f$ is uniformly continuous, you can take $deltain (2^{-(n+1)},2^{-n})$ and convolute $f+delta$ with a mollifier $eta_epsilon$. The resulting function is smooth and converges uniformly to $f+delta$ as $epsilonsearrow 0$. One can probably make something similar work with the assumption of uniform continuity.
$endgroup$
– MaoWao
Jan 25 at 11:30
add a comment |
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$begingroup$
If $f$ is uniformly continuous, you can take $deltain (2^{-(n+1)},2^{-n})$ and convolute $f+delta$ with a mollifier $eta_epsilon$. The resulting function is smooth and converges uniformly to $f+delta$ as $epsilonsearrow 0$. One can probably make something similar work with the assumption of uniform continuity.
$endgroup$
– MaoWao
Jan 25 at 11:30