Uniformly continuous functions in $mathbb{C}$ over paths and term by term integration
Theorem.
Let $gamma$ be a path defined on $[a,b]$ in an open set $U$ and let $g$ be a continuous function on $gamma$ (that is, on the image $gamma( [a,b] )$). If $z$ is not on $gamma$, define $$f(z) = int _{gamma} frac{g(zeta )}{zeta - z} d zeta .$$ Then $f$ is analytic on the complement of $gamma$ in $U$, and its derivatives are given by
$$f^{(n)}(z) = n! int _{gamma} frac{g(zeta )}{(zeta - z)^{n+1}} d zeta .$$
Partial Proof.
Let $z_0 in U$ and $z_0$ not on $gamma$. Since the image of $gamma$ is compact, there exists a minimum distance between $z_0$ and points on $gamma$. Choose $0<R<d(z_0 , gamma)$ and take $R$ small enough so that the closed disc $overline{D} (z_0 , R)$ is contained in $U$. Take $0<s<R$ and replace $f$ by $g$ inside the integral sign. Expand $1/(zeta - z)$ by geometric series and proceed to see that $f$ has a power series expansion…
I understand the rest of the proof, but the bold is where I am struggling. Filling in the details by expanding using geometric series, we can see that the expansion converges absolutely and uniformly for $|z-z_0| leq s$.
However, I cannot understand the wording of my instructor: $g$ is continuous on $gamma$, so it is bounded and therefore we can apply the following theorem to integrate term by term:
Theorem.
Let ${f_n}$ be a sequence of continuous functions on $U$ which converge uniformly to $f$. Then $$lim int _{gamma} f_n = int _{gamma} f.$$ If $sum f_n$ is a series of continuous functions which converge uniformly on $U$ then $$ int _{gamma} sum f_n = sum int _{gamma} f_n.$$
I have two questions: to use the theorem, $g$ needs to be uniformly continuous. I know that $g$ is uniformly continuous if it is continuous on a compact set. So, how do we know $g$ is uniformly continuous on the path? I think we are assuming again in the problem that $g$ is continuous on the image of $gamma$ (which is compact), right? The second question is: if this is the case, why only mention that $g$ is continuous, which implies bounded, rather than just stating $g$ is continuous on the image, which is compact, which implies uniformly continuous? (Are there details that are stopping us from concluding $g$ is uniformly continuous right away, such as having to use the boundedness condition to deduce uniformly continuous since we are on the image of a path?)
complex-analysis
asked Nov 28 '18 at 17:56
CompactCompact
1818
add a comment |
Theorem.
Let $gamma$ be a path defined on $[a,b]$ in an open set $U$ and let $g$ be a continuous function on $gamma$ (that is, on the image $gamma( [a,b] )$). If $z$ is not on $gamma$, define $$f(z) = int _{gamma} frac{g(zeta )}{zeta - z} d zeta .$$ Then $f$ is analytic on the complement of $gamma$ in $U$, and its derivatives are given by
$$f^{(n)}(z) = n! int _{gamma} frac{g(zeta )}{(zeta - z)^{n+1}} d zeta .$$
Partial Proof.
Let $z_0 in U$ and $z_0$ not on $gamma$. Since the image of $gamma$ is compact, there exists a minimum distance between $z_0$ and points on $gamma$. Choose $0<R<d(z_0 , gamma)$ and take $R$ small enough so that the closed disc $overline{D} (z_0 , R)$ is contained in $U$. Take $0<s<R$ and replace $f$ by $g$ inside the integral sign. Expand $1/(zeta - z)$ by geometric series and proceed to see that $f$ has a power series expansion…
I understand the rest of the proof, but the bold is where I am struggling. Filling in the details by expanding using geometric series, we can see that the expansion converges absolutely and uniformly for $|z-z_0| leq s$.
However, I cannot understand the wording of my instructor: $g$ is continuous on $gamma$, so it is bounded and therefore we can apply the following theorem to integrate term by term:
Theorem.
Let ${f_n}$ be a sequence of continuous functions on $U$ which converge uniformly to $f$. Then $$lim int _{gamma} f_n = int _{gamma} f.$$ If $sum f_n$ is a series of continuous functions which converge uniformly on $U$ then $$ int _{gamma} sum f_n = sum int _{gamma} f_n.$$
I have two questions: to use the theorem, $g$ needs to be uniformly continuous. I know that $g$ is uniformly continuous if it is continuous on a compact set. So, how do we know $g$ is uniformly continuous on the path? I think we are assuming again in the problem that $g$ is continuous on the image of $gamma$ (which is compact), right? The second question is: if this is the case, why only mention that $g$ is continuous, which implies bounded, rather than just stating $g$ is continuous on the image, which is compact, which implies uniformly continuous? (Are there details that are stopping us from concluding $g$ is uniformly continuous right away, such as having to use the boundedness condition to deduce uniformly continuous since we are on the image of a path?)
complex-analysis
asked Nov 28 '18 at 17:56
CompactCompact
1818
We are not assuming that $g$ is continuous on $gamma$. It says so right in the statement of the theorem. I suspect your instructor was thinking of "closed and bounded implies compact" instead of "continuous image of compact is also compact", because that is how your instructor thinks. We all have our own way of approaching things, and it isn't always the most effecient.
– Paul Sinclair
Nov 29 '18 at 0:35
add a comment |
Theorem.
Let $gamma$ be a path defined on $[a,b]$ in an open set $U$ and let $g$ be a continuous function on $gamma$ (that is, on the image $gamma( [a,b] )$). If $z$ is not on $gamma$, define $$f(z) = int _{gamma} frac{g(zeta )}{zeta - z} d zeta .$$ Then $f$ is analytic on the complement of $gamma$ in $U$, and its derivatives are given by
$$f^{(n)}(z) = n! int _{gamma} frac{g(zeta )}{(zeta - z)^{n+1}} d zeta .$$
Partial Proof.
Let $z_0 in U$ and $z_0$ not on $gamma$. Since the image of $gamma$ is compact, there exists a minimum distance between $z_0$ and points on $gamma$. Choose $0<R<d(z_0 , gamma)$ and take $R$ small enough so that the closed disc $overline{D} (z_0 , R)$ is contained in $U$. Take $0<s<R$ and replace $f$ by $g$ inside the integral sign. Expand $1/(zeta - z)$ by geometric series and proceed to see that $f$ has a power series expansion…
I understand the rest of the proof, but the bold is where I am struggling. Filling in the details by expanding using geometric series, we can see that the expansion converges absolutely and uniformly for $|z-z_0| leq s$.
However, I cannot understand the wording of my instructor: $g$ is continuous on $gamma$, so it is bounded and therefore we can apply the following theorem to integrate term by term:
Theorem.
Let ${f_n}$ be a sequence of continuous functions on $U$ which converge uniformly to $f$. Then $$lim int _{gamma} f_n = int _{gamma} f.$$ If $sum f_n$ is a series of continuous functions which converge uniformly on $U$ then $$ int _{gamma} sum f_n = sum int _{gamma} f_n.$$
I have two questions: to use the theorem, $g$ needs to be uniformly continuous. I know that $g$ is uniformly continuous if it is continuous on a compact set. So, how do we know $g$ is uniformly continuous on the path? I think we are assuming again in the problem that $g$ is continuous on the image of $gamma$ (which is compact), right? The second question is: if this is the case, why only mention that $g$ is continuous, which implies bounded, rather than just stating $g$ is continuous on the image, which is compact, which implies uniformly continuous? (Are there details that are stopping us from concluding $g$ is uniformly continuous right away, such as having to use the boundedness condition to deduce uniformly continuous since we are on the image of a path?)
complex-analysis
asked Nov 28 '18 at 17:56
CompactCompact
1818
Theorem.
Let $gamma$ be a path defined on $[a,b]$ in an open set $U$ and let $g$ be a continuous function on $gamma$ (that is, on the image $gamma( [a,b] )$). If $z$ is not on $gamma$, define $$f(z) = int _{gamma} frac{g(zeta )}{zeta - z} d zeta .$$ Then $f$ is analytic on the complement of $gamma$ in $U$, and its derivatives are given by
$$f^{(n)}(z) = n! int _{gamma} frac{g(zeta )}{(zeta - z)^{n+1}} d zeta .$$
Partial Proof.
Let $z_0 in U$ and $z_0$ not on $gamma$. Since the image of $gamma$ is compact, there exists a minimum distance between $z_0$ and points on $gamma$. Choose $0<R<d(z_0 , gamma)$ and take $R$ small enough so that the closed disc $overline{D} (z_0 , R)$ is contained in $U$. Take $0<s<R$ and replace $f$ by $g$ inside the integral sign. Expand $1/(zeta - z)$ by geometric series and proceed to see that $f$ has a power series expansion…
I understand the rest of the proof, but the bold is where I am struggling. Filling in the details by expanding using geometric series, we can see that the expansion converges absolutely and uniformly for $|z-z_0| leq s$.
However, I cannot understand the wording of my instructor: $g$ is continuous on $gamma$, so it is bounded and therefore we can apply the following theorem to integrate term by term:
Theorem.
Let ${f_n}$ be a sequence of continuous functions on $U$ which converge uniformly to $f$. Then $$lim int _{gamma} f_n = int _{gamma} f.$$ If $sum f_n$ is a series of continuous functions which converge uniformly on $U$ then $$ int _{gamma} sum f_n = sum int _{gamma} f_n.$$
I have two questions: to use the theorem, $g$ needs to be uniformly continuous. I know that $g$ is uniformly continuous if it is continuous on a compact set. So, how do we know $g$ is uniformly continuous on the path? I think we are assuming again in the problem that $g$ is continuous on the image of $gamma$ (which is compact), right? The second question is: if this is the case, why only mention that $g$ is continuous, which implies bounded, rather than just stating $g$ is continuous on the image, which is compact, which implies uniformly continuous? (Are there details that are stopping us from concluding $g$ is uniformly continuous right away, such as having to use the boundedness condition to deduce uniformly continuous since we are on the image of a path?)
complex-analysis
complex-analysis
asked Nov 28 '18 at 17:56
CompactCompact
1818
asked Nov 28 '18 at 17:56
CompactCompact
1818
asked Nov 28 '18 at 17:56
CompactCompact
1818
asked Nov 28 '18 at 17:56
CompactCompact
1818
asked Nov 28 '18 at 17:56
CompactCompact
1818
1818
We are not assuming that $g$ is continuous on $gamma$. It says so right in the statement of the theorem. I suspect your instructor was thinking of "closed and bounded implies compact" instead of "continuous image of compact is also compact", because that is how your instructor thinks. We all have our own way of approaching things, and it isn't always the most effecient.
– Paul Sinclair
Nov 29 '18 at 0:35
add a comment |
We are not assuming that $g$ is continuous on $gamma$. It says so right in the statement of the theorem. I suspect your instructor was thinking of "closed and bounded implies compact" instead of "continuous image of compact is also compact", because that is how your instructor thinks. We all have our own way of approaching things, and it isn't always the most effecient.
– Paul Sinclair
Nov 29 '18 at 0:35
We are not assuming that $g$ is continuous on $gamma$. It says so right in the statement of the theorem. I suspect your instructor was thinking of "closed and bounded implies compact" instead of "continuous image of compact is also compact", because that is how your instructor thinks. We all have our own way of approaching things, and it isn't always the most effecient.
– Paul Sinclair
Nov 29 '18 at 0:35
We are not assuming that $g$ is continuous on $gamma$. It says so right in the statement of the theorem. I suspect your instructor was thinking of "closed and bounded implies compact" instead of "continuous image of compact is also compact", because that is how your instructor thinks. We all have our own way of approaching things, and it isn't always the most effecient.
– Paul Sinclair
Nov 29 '18 at 0:35
add a comment |
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We are not assuming that $g$ is continuous on $gamma$. It says so right in the statement of the theorem. I suspect your instructor was thinking of "closed and bounded implies compact" instead of "continuous image of compact is also compact", because that is how your instructor thinks. We all have our own way of approaching things, and it isn't always the most effecient.
– Paul Sinclair
Nov 29 '18 at 0:35