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I started working on a problema on building a sequence of continuous function whose pointwise limit has to be a real valued function f. I mean: $f:R^2 \rightarrow [-\infty; +\infty]$ and I know that the two function fixing the single variables defined as:
h:$t \rightarrow f(x,t)$
g:$t \rightarrow f(t,y)$
are continuous. I'm asked to prove that f is the pointwise limit of a sequence of continuous function (and since it is proved it also measurable).

So I started working with the variables y fixed;imaging that if I consider f for each $(x,y+\frac{1}{2^n})$ the succesion $f_n=f(x,y+\frac{1}{2^n})$.
I can get that this tends to f (using the conitinuity of g) but $f_n$ as defined is not continuous.So I don't know how to solve this problem.May I use some difference between two points as continuous function (where the continuity is given by the functions g and h)? I'm in lack if ideas..

1 Answers 1

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You may construct $f_n$ as follows.

$$f_n(x,y)=(k+1-2^ny)f(x,\frac{k}{2^n})+(2^ny-k)f(x,\frac{k+1}{2^n}),\quad \frac{k}{2^n}\le y<\frac{k+1}{2^n}, k\in\mathbb{Z}.$$

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    May I consider this functions as the function passing through the points $f(x,\frac{k}{2^n})$ and $f(x,\frac{k+1}{2^n})$? When $n \rightarrow \infty$ both this point and y comes to 0.so since x is fixed and g is continuous all the points $(x,y),(x,\frac{k}{2^n}),(x,\frac{k+1}{2^n})$ tends to zero and the extreme of the interval to y and so all the images. $\vert f_n -f(x,y)\vert= \vert k[f(x,\frac{k}{2^n})-f(x,\frac{k+1}{2^n})] + 2^n*y[f(x,\frac{k+1}{2^n})-f(x,\frac{k}{2^n})]+ f(x,\frac{k}{2^n})-f(x,y)\vert$ I can share into three part,all of that converges because of g continuity.Right?2012-11-20
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    @Laura: Your decomposition of $|f(x,y)-f_n(x,y)|$ is not good.2012-11-20
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    @Laura: Given $y$ and $n$, there exists $k_n\in\mathbb{Z}$, such that $\frac{k_n}{2^n}\le y< \frac{k_n+1}{2^n}$. Denote $\lambda=2^ny-k_n$. Note that $0\le\lambda<1$, and $f_n(x,y)=(1-\lambda)f(x,\frac{k_n}{2^n})+\lambda f(x,\frac{k_n+1}{2^n})$. This helps you to find a better decomposition of $|f(x,y)-f_n(x,y)|$.2012-11-20
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    @Laura: Are you clear now?2012-11-20
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    My decompositions above is messed up: from the first comment:$\vert k\vert\vert f(x,\frac{k+1}{2^n})-f(x,\frac{k}{2^n})+\vert 2^n*y \vert \vert f(x,\frac{k}{2^n})-f(x,\frac{k+1}{2^n})\vert + \vert f(x,y)-f(x,\frac{k}{2^n})$ From the second: $\vert f(x,y)-f(x,\frac{k_n}{2^n})\vert + \vert \lambda \vert \vert f(x,\frac{k_n+1}{2^n})-f(x,\frac{k_n}{2^n})\vert $ What's wrong in the first one?2012-11-20
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    @Laura: How can you show the first one tends to $0$ when $n\to\infty$?2012-11-20
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    k is fixed, $\vert f(x,\frac{k+1}{2^n})-f(x,\frac{k}{2^n})\vert $since g is continuous and $\frac{k+1}{2^n};\frac{k}{2^n}$ tends to y so do the images. I know $2^n$ tends to $\infty$ but $y$ tends to $0$ at the same time.I think that's all..is basically the rules which stands above your second suggestion but less "elegant".2012-11-20
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    @Laura: $y$ is fixed at the beginning, and $k$ is not fixed.2012-11-20
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    Ok,while in your second suggestion it is.. Sorry I confused the two.But in the second $k_n$ is fixed to have $\lambda$ as described.2012-11-20
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    @Laura: $k_n$ depends on both $y$ and $n$. So does $\lambda$. However, it does not matter, because $0\le\lambda<1$.2012-11-20
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    Yes,that was exactly what I meant to tell,I wasn't not clear but I got it.Thank you.2012-11-20
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    @Laura: You are welcome!2012-11-20
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    Just a last thing: can I assume that $f_n=(1-\lambda)f(x,\frac{k}{2^n})+\lambda f(x,\frac{k+1}{2^n})$ is continuous just considering it as a sum and products of a polinom (because of y in $\lambda$) and constant?Since k,n are fixed for every y.2012-11-21
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    @Laura: It is not accurate to say "$k$, $n$ are fixed for every $y$", because if $y$ and $n$ are given first, the choice of $k$ depends on $n$ and $y$. Nevertheless, if $k$, $n$ are given first, on the interval $[\frac{k}{2^n},\frac{k+1}{2^n}]$, as you have seen, $f_n$ is a sum of two terms, and both terms are product of two continuous functions, so $f_n$ is continuous.2012-11-21