Is this an example of a pointwise convergent sequence of function but not uniformly convergent

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Define a function $f_n(x) = |x-2|$ for $x \in (n-2, n+2)$ and zero everywhere.

Then seems to me for each x, for a sufficiently large n $f_n(x)$ approaches to zero. but it does not converge uniformly because we will always have the "bump" for some n.

Furthermore the integral of $f_n(x)$ does not converge to the integral of the pointwise limit $f = 0$. since the integral of $f_n(x)$ is the same for all $n$.

Now part (b) of the question ask if it is possible to come up with an example of $f_n(x)$ such that the limit of the integral of $f_n$ converges to the integral of the pointwise limit, but $f_n(x)$ does not converge uniformly to the limit.

I am thinking I can constructing something that converges to the dirac delta function at 0, and a negative bum that goes off the infinity with equal area of the dirac delta "bump" in magnitude. But this seems problematic since there is an infinity at zero so I can't really say the integral of the pointwise limit is 0.

Would it work if I have two dirac delta "bumps" that tends off to positive and negative infinity as n -> infinity?

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Part a has been addressed in the comments, the way you have formulated things in the question is wrong but for somewhat trivial reasons.

To address part b, a better sequence would be to take,

$$\forall n\in \mathbb{Z}: f_n(x) = \begin{cases}x-n & x\in(n-1,n+1)\\ 0 & x\not\in(n-1,n+1) \end{cases}$$

Then, $\int_{\mathbb{R}}f_n(x)dx = 0$ for all $n\in\mathbb{Z}$. Furthermore, the pointwise limit of $f_n$ is $f\equiv 0$ for all $x$, which has integral $\int_{\mathbb{R}}f(x)dx=0$, no need for delta functions.