2

Suppose $f_n\to f$ almost everywhere for all $x\in X=(0,1)$ and for some fixed $M<\infty$, we have: $$\sup_n |{f_n}|_{L^2(X)}\leq M.$$

(a) Is it true that $\lim_{n\to \infty}|{f_n- f}|_{L^2(X)}=0$? If so, prove it.

(b) Now, suppose that $\lim_{n\to \infty}|{f_n}|_{L^2(X)} = |{f}|_{L^2(X)}$. Does the limit in (a) hold?

I think part (a) is false, with the following counter example: Let $\{f_n\}$ on $X = (0, 1)$ be a sequence of functions such that each $f_n$ is a moving bump that gets narrower and taller as $n$ increases but always integrates to 1 (I think this is the Dirac Deta function). This sequence can converge almost everywhere to the zero function (since each point is eventually outside the support of all bumps), but the $L^2$ norm of $f_n - f$ where $f$ is the zero function will not converge to 0. This is because the integral of the square of each bump does not vanish despite the bumps becoming arbitrarily narrow.

I'm not too sure about part (b) and how it would affect the answer for (a).

Feng
  • 13,705
Hector Lombard
  • 649
  • 3
  • 11

1 Answers1

1

For a), an explicit counter-example can be $f_n(x)=\sqrt n\mathbf{1}_{(0,1/n)}(x)$. For each $x\in (0,1)$, $f_n(x)$ vanishes for $n$ larger than say $1+1/x$ hence $f_n\to 0=f$ almost everywhere but $\lVert f_n -f\rVert_2=1$.

For b), this is a special case of this thread.

Davide Giraudo
  • 172,925