Brezis' exercise 4.19

Question

I'm having trouble with problem 4.19 of Brezis' book in functional analysis (Functional Analysis, Sobolev Spaces and Partial Differential Equations). It asks for a sequence $f_n\geq 0$ in $L^1(0,1)$ and a function $f\in L^1(0,1)$ such that

1. $f_n \to f$ weakly $\sigma(L^1,L^\infty)$;
2. $||f_n||_1 \to ||f||_1$;
3. $||f_n-f||_1 \not \to 0$.

What sequence $f_n$ work? I know that i cannot choose $f=0$ because statement 2 would contradicts 3.

Answer 1

Consider $r_n \colon \mathbb{R} \to \mathbb{R}$ given by $r_0(x) = (-1)^{\lfloor x\rfloor}$ and $r_n(x) = r_0(2^n\cdot x)$. Let $g_n = r_n\lvert_{(0,1)}$. Then show that $g_n \to 0$ weakly, and use these functions to construct the desired sequence $(f_n)$.

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Accepting that $g_n \to 0$ weakly for the moment, one sees that for every $f\in L^1(0,1)$ with $f(x) \geqslant 1$ for all $x$, by setting $f_n = f + g_n$ we have a sequence $f_n \geqslant 0$ in $L^1(0,1)$ with $f_n \to f$ weakly, and $\lVert f_n - f\rVert_1 = \lVert g_n\rVert_1 = 1$ for all $n$. For $n \geqslant 1$, we also have

$$\lVert f_n\rVert_1 = \int_0^1 f_n(x)\,dx = \int_0^1 f(x)\,dx + \int_0^1 g_n(x)\,dx = \int_0^1 f(x)\,dx = \lVert f\rVert_1.$$

So it remains to see that $g_n \to 0$ weakly. First we show that

$$\int_0^1 g_n(x)h(x)\,dx \to 0\tag{1}$$

for all $h \in C([0,1])$. Since $[0,1]$ is compact, every $h\in C([0,1])$ is uniformly continuous. Given $\varepsilon > 0$, we can therefore find a $\delta > 0$ such that $\lvert x-y\rvert \leqslant \delta \implies \lvert h(x) - h(y)\rvert \leqslant \varepsilon$. Then let $n \geqslant 1$ such that $2^{-n} \leqslant \delta$. For $0 < k < 2^n$ we have

\begin{align} \biggl\lvert\int_{2^{-n}(k-1)}^{2^{-n}(k+1)} g_n(x)h(x)\,dx\bigr\rvert &= \biggl\lvert\int_{2^{-n}(k-1)}^{2^{-n}(k+1)}g_n(x)\bigl(h(x) - h(2^{-n}k)\bigr)\,dx\bigr\rvert\\ &\leqslant \int_{2^{-n}(k-1)}^{2^{-n}(k+1)} \lvert g_n(x)\rvert\cdot\lvert h(x) - h(2^{-n}k)\rvert\,dx\\ &\leqslant 2^{1-n}\varepsilon. \end{align}

Summing over all odd $k$ between $0$ and $2^n$, we obtain

$$\biggl\lvert \int_0^1 g_n(x)h(x)\,dx\biggr\rvert \leqslant \varepsilon.$$

Thus $(1)$ is proved.

Next we take an arbitrary $h \in L^\infty(0,1)$. Given $\varepsilon > 0$, by Luzin's theorem there is a $h_\varepsilon \in C([0,1])$ with $\lVert h_\varepsilon\rVert_\infty \leqslant \lVert h\rVert_\infty$ such that $\lambda(\{ x : h_\varepsilon(x) \neq h(x)\}) < \varepsilon$. Then we have

\begin{align} \biggl\lvert \int_0^1 g_n(x)h(x)\,dx\biggr\rvert &\leqslant \biggl\lvert \int_0^1 g_n(x)h_\varepsilon(x)\,dx\biggr\rvert + \biggl\lvert \int_0^1 g_n(x)\bigl(h(x) - h_\varepsilon(x)\bigr)\,dx\biggr\rvert\\ &\leqslant \biggl\lvert \int_0^1 g_n(x)h_\varepsilon(x)\,dx\biggr\rvert + \biggl\lvert \int_{\{ x : h_\varepsilon(x) \neq h(x)\}} g_n(x)\bigl(h(x) - h_\varepsilon(x)\bigr)\,dx\biggr\rvert\\ &\leqslant \biggl\lvert \int_0^1 g_n(x)h_\varepsilon(x)\,dx\biggr\rvert + \int_{\{ x : h_\varepsilon(x) \neq h(x)\}}\lvert h(x) - h_\varepsilon(x)\rvert\,dx\\ &\leqslant \biggl\lvert \int_0^1 g_n(x)h_\varepsilon(x)\,dx\biggr\rvert + 2\lVert h\rVert_\infty \varepsilon, \end{align}

and hence

$$\limsup_{n\to\infty}\: \biggl\lvert \int_0^1 g_n(x)h(x)\,dx\biggr\rvert \leqslant 2\lVert h\rVert_\infty \varepsilon.$$

Since that holds for all $\varepsilon > 0$, we conclude

$$\lim_{n\to\infty} \int_0^1 g_n(x) h(x)\,dx = 0,$$

as desired.