Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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..

share|cite|improve this question
up vote 0 down vote accepted

You may construct $f_n$ as follows.

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

share|cite|improve this answer
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 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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.