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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..

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1 Answer 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? –  Laura Nov 20 '12 at 15:12
    
@Laura: Your decomposition of $|f(x,y)-f_n(x,y)|$ is not good. –  23rd Nov 20 '12 at 15:23
    
@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 $|f(x,y)-f_n(x,y)|$. –  23rd Nov 20 '12 at 15:31
    
@Laura: Are you clear now? –  23rd Nov 20 '12 at 15:44
    
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? –  Laura Nov 20 '12 at 15:44
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