I still want to examine different convergence results of the sequence of functions $f_n(x)= e^{-\frac{x^2}{n^2}}$ on $\mathbb{R}$. Convergence in the sense of distributions is shown in the following post: Convergence in $\mathcal{D}'(\mathbb{R})$ . Now I want to check the convergence in Convergence in $L^p(\mathbb{R})$ for $1\leq p \leq \infty$. How would you start the analysis? Guess for a limit (in this case $1$) and then compute $\vert\vert f_n -1\vert \vert_{L^p(\mathbb{R})}$? Or is there another trick or theorem with an equivalent condition...?


There is a problem. The constant function $1$ is not in $L^p(\Bbb R)$ if $p\in[1,\infty)$, and $\|f_n-1\|_p=\infty$ for all $n$.

If $p=\infty$, $f_n$ converges pointwise to $1$, but not in the $L^\infty$ norm, since$\|f_n-1\|_\infty=1$ for all $n$.

  • $\begingroup$ sorry, I am a little confused now. When I am asked to say whether a sequence of functions converge in $L^p(\mathbb{R})$, then I analyze the norm as above? Or what is the difference you mentioned above with the pointwise limit? $\endgroup$
    – tubmaster
    Mar 20 '17 at 14:33
  • $\begingroup$ A sequence $\{z_n\}$ converges to $z$ in a normed space $X$ if $\lim_{n\to0}\|z_n-z\|_X=0$ (which in particular implies that $\{x_n\}$ is bounded in $X$.) The sequence $f_n$ does not converge in $L^p$. You can check that $\|f_n\|_p$ is unbounded if $1\le p<\infty$. In the case $p=\infty$, the sequence does not converge in $L^\infty$; the only possible limit would be the constant function $1$, but $\|f_n-1\|_\infty$ does not converge to $0$. $\endgroup$ Mar 20 '17 at 14:41
  • $\begingroup$ Why is $\|f_n\|_p$ unbounded? $\endgroup$
    – tubmaster
    Mar 20 '17 at 14:47
  • $\begingroup$ Calculate it. Hint: do the change of variable $x=n\,t$. $\endgroup$ Mar 20 '17 at 14:49
  • $\begingroup$ ah okay I see ... after change of variable there comes a $n$ in front of the integral, right? $\endgroup$
    – tubmaster
    Mar 20 '17 at 15:08

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