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I have this continuous function $f:\mathbb{C}\rightarrow\mathbb{C}$ defined on an open set $\Omega$.
I also have a family of identical smooth curves up to translation $z_{t}:[a,b]\rightarrow\mathbb{C}$ in $\Omega$ and $t \in [0, T]$, where the parameter $t$ specifies some linear translation, i.e. $z_{t+\delta} = z_{t} + \delta K$ where $K$ is some constant. (You can imagine that the curve $z_{t}$ is "shifting" with the passage of time, $t$.)

So my question is, is the integral $\int_{z_{t}}f(x) dx$ continuous as a function of $t$? In other words, as my curve $z_{t}$ moves a little bit in $\Omega$, does the value of the integral also move a little bit?

More specifically, in my case I know that the value of the integral is $0$ for all $t \in (\alpha, \beta]$. Must the value of the integral also be zero for $t=\alpha$?

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Write down the definitions: $$\int_{z_t} f(x)dx = \int_a^b f(z_t(s))\frac{d}{ ds} z_t(s) ds$$ With your 'translation' property of the $z_t$ the derivative wrt $s$ does not depend on $t$, so only the $f(z_t(s))$ plays a role. Now write down the difference for $t = \alpha$ and $t$ close to $\alpha$ and estimate the difference: $$\int_a^b \left(f(z_\alpha(s)) - f(z_t(s))\right)\frac{d}{ ds} z_t(s) ds $$ Since the image of $[0,T]\times [a,b]$ under $z$ is compact und $f$ continuous, $f$ is uniformly continuous on that image, hence the difference can be made uniformly arbitrarily small as $t\rightarrow \alpha$, given your definition of $z_t$.

(This reasoning, of course, relies on the assumption that the images of all $z_t$ are contained in $\Omega$ and do not, e.g. tend to the boundary of $\Omega$ as $t\rightarrow \alpha$. But since you've chosen closed intervals as domain of definition, I assume this to be true.)

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Sorry, I don't quite understand the proof. Is it possible to make it more rigorous? –  Mark Jun 4 '12 at 16:30
    
@Mark Which part is not rigorous enough for you? –  user20266 Jun 4 '12 at 16:40
    
Let $F(t,s) = f(z_{t}(s))$. Then $F:[0,T]\times[a,b]\rightarrow\mathbb{C}$ It seems reasonable that $F$ is continuous on its domain and therefore F is uniformly continuous. Now choose an $\epsilon$ and we get $|F(\alpha, s) - F(0, s)| < \epsilon$ when we choose $\alpha < \delta$. Then $\int_{a}^{b}|F(\alpha,s)-F(0,s)|z'(s)ds < \epsilon\int_{a}^{b}z'(s)ds$. Since $\epsilon$ is arbitrary, the integral is $0$. Is this correct? –  Mark Jun 4 '12 at 17:05
    
@Mark Kind of. I think you mixed up $\alpha, t$ and $0$ when compared with your posting. $\alpha$ already has a special meaning in your original question. Then, first of all, yes, your $F$ is continuous as composition of continuous maps. Then choose $\epsilon$ such that $|F(\alpha,s) - F(t,s)|\le \epsilon$ for $|t-\alpha|< \epsilon$. Then replace $0$ in the integral by $t$, and in order to estimate the (norm of the) integrals you want to take the norm of the derivative of $z_t$ wrt to $s$ otherwise you might get in trouble with the signs. Or replace $\alpha, \beta$ in your OP by $0,T$. –  user20266 Jun 4 '12 at 17:28
    
Yes, of course, I forgot to norm the derivatives. Thanks a bunch! If I fix the errors, your comment will no longer make sense so I guess I'll leave it. –  Mark Jun 4 '12 at 18:22
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