If $f_n\to f$ weakly in $L^2(a,b)$ and $f_n\to g$ strongly in $L^1(a,b)$ do me necessarily have $f=g$?

Question

1) If $f_n\to f$ weakly in $L^2(a,b)$ and $f_n\to g$ strongly in $L^1(a,b)$ do me necessarily have $f=g$ ?

2) More generally, if $\Omega \subset \mathbb R^n$ bounded, open with Lipschitz boundary, do we have that $f_n\to f$ strongly in $L^p(\Omega )$ and $f_n\to g$ weakly in $L^q(\Omega )$, then $f=g$ ? Here we suppose that $f_n,f,g$ are in $L^p(\Omega )$ and $L^q(\Omega )$ (but we'll distinguish the cases $p<q$ and $p>q$).

Attempt

1) What I observe is $1\in L^1(a,b)$ and thus $$\lim_{n\to \infty }\int_a^b f_n=\int_a^b f,$$ but I know that doesn't mean that $\lim_{n\to \infty }\int |f_n-f|=0$, and I can't prove it since $$\int_a^b |f_n-f|\geq \left|\int_a^b (f_n-f)\right|.$$ So is it true ?

2) No idea.

Answer 1

Your conditions (for both your questions) imply in particular that $\langle f-g,\phi\rangle = 0$ for all $\phi\in C^{\infty}_c$ hence $f=g$ a.e. by the fundamental lemma of calculus of variations.

More details: Denoting $\langle f, g\rangle:=\int f g.$ Recall $f_n\to f$ weakly in $L^2$ means $\langle f_n-f,\phi\rangle\to0$ for all $\phi\in L^2$ (in particular for $\phi \in C_c^{\infty}$). And if $f_n\to g$ in $L^1$, then by Holder's inequality $\langle f_n -g, \phi\rangle\leq \|f_n-g\|_{L^1}\|\phi\|_{\infty}\to 0$ as $n\to \infty$. Combining these, $\langle f-g,\phi\rangle=0$ for all $\phi\in C_c^{\infty}$.

Answer 2

This works in a more general context.

Let $\left(X,\mathcal A,\mu\right)$ be a $\sigma$-finite measure space and let $1\leqslant p,q<\infty$. Assume that $\left(f_n\right)_{n\geqslant 1}$ is a sequence of elements of $\mathbb L^q\cap \mathbb L^p$ which convergence weakly to $f$ in $\mathbb L^p$ and strongly to $g$ in $\mathbb L^q$. Then $f=g$ almost everywhere.

Indeed, for any $A\in\mathcal A$ of finite measure, the indicator function of $A$ belongs to the dual space of $\mathbb L^p$ hence $$\tag{1}\lim_{n\to +\infty}\int f_n\mathbf 1_A\mathrm d\mu=\int f\mathbf 1_A\mathrm d\mu.$$ Using Hölder's inequality, it follows that $$\tag{2}\lim_{n\to +\infty}\int f_n\mathbf 1_A\mathrm d\mu=\int g\mathbf 1_A\mathrm d\mu.$$ The combination of (1) and (2) yields $$\tag{3}\int \left(f-g\right)\mathbf 1_A\mathrm d\mu=0.$$ Let $F\in\mathcal A$ be a set of finite measure. Using (3) with $A=F\cap\left\{f-g\geqslant 1/k\right\}$, we derive that $\mu\left(F\cap\left\{f-g\geqslant 1/k\right\}\right)=0$ for any integer $k$ hence $\mu\left(F\cap\left\{f-g\gt 0\right\}\right)=0$. A similar reasoning gives $\mu\left(F\cap\left\{g-f\gt 0\right\}\right)=0$ hence $$\tag{4}\mu\left(F\cap\left\{f-g\neq 0\right\}\right)=0.$$ Since $\left(X,\mathcal A,\mu\right)$ is supposed to be $\sigma$-finite, we can write $X$ as a countable union $\bigcup_{n\geqslant 1}F_n$ of elements of $\mathcal A$ of finite measure. Using (4) with $F=F_n$, we derive that the measure of $\left\{f-g\neq 0\right\}$ is zero.

Answer 3

Although I like the answer given by Davide Giraudo, let me give you another proof of the statement using Egorov's theorem. We'd like to prove the following statement:

Statement 1: Let $\Omega \subset \mathbb{R}^n$ be an open set and let $1 \leq p, q < \infty$ be given. Assume that $g\in L^q(\Omega)$, $f \in L^p(\Omega)$, and $f_n \in L^p (\Omega) \cap L^q(\Omega)$ for all $n \in \mathbb{N}$. If $f_n \rightharpoonup g$ in $L^q(\Omega)$ and $f_n \rightarrow f$ in $L^p(\Omega)$, then $f = g$ almost everywhere.

The above statement is a consequence of the following one:

Statement 2: Let $(X, \mathcal{A}, \mu)$ be a $\sigma$-finite measure space and let $1 \leq p, q < \infty$. Assume that $g\in L^q(X, \mu)$, $f \in L^p(X, \mu)$, and $f_n \in L^q(X, \mu)$ for all $n \in \mathbb{N}$. If $f_n\rightharpoonup g$ in $L^q(X, \mu)$ and $f_n \to f$ almost everywhere in $X$, then $f = g$ almost everywhere in $X$.

All functions are assumed to be real-valued.

Proof of Statement 2: Step 1: Let us first assume that $(X, \mathcal{A}, \mu)$ is a finite measure space. By Egorov's theorem, for every $\varepsilon > 0$ we find a measurable (good) subsset $G_{\varepsilon} \subset X$ such that \begin{equation} \mu(X \setminus G_{\varepsilon}) < \varepsilon \quad \text{ and } \quad \lim_{n \to \infty} \| f - f_n \|_{L^\infty(G_{\varepsilon})} = 0\, . \end{equation} Now, fix $\varphi \in C_c^{\infty}(X)$. Using the triangle inequality, we find that \begin{equation} \left| \int_{X} (f-g) \varphi \right| \leq \int_{X} \lvert f-f_n \rvert \lvert \varphi \rvert + \left| \int_{X} (f_n - g) \varphi \right|\, . \end{equation} The second integral on the right-hand side tends to $0$ as $n \to \infty$, because $f_n \rightharpoonup g$ in $L^q(\Omega)$. The first integral can be estimated as follows \begin{align} \int_{X} \lvert f - f_n \rvert \lvert \varphi\rvert &= \int_{X \setminus G_{\varepsilon}} \lvert f-f_n\rvert \lvert \varphi \rvert + \int_{G_{\varepsilon}} \lvert f - f_n \rvert \lvert \varphi \rvert \\ &\leq \varepsilon \Big(| f \|_{L^p(X, \mu)} \| \varphi^{p'} \|_{\infty} + \| f_n \|_{L^q(X, \mu)} \| \varphi^{q'}\|_{\infty} \Big) + \| f -f_n \|_{L^{\infty}(G_{\varepsilon})} \| \varphi \|_{L^1(X)}\, . \end{align} Note that the sequence $(\| f_n \|_{L^q(X, \mu)})_n$ is bounded since $(f_n)_n$ converges weakly to $g$ in $L^q(X, \mu)$. Thus, sending $n \to \infty$, we find that for all $\varepsilon > 0$ \begin{equation} \left| \int_{X} (f-g) \varphi \right| \leq C(f, \varphi) \varepsilon\, . \end{equation} As $\varphi$ was chosen arbitrarily, we conclude that $f = g$ almost everywhere in $X$.

Step 2: Let us now assume that $(X, \mathcal{A}, \mu)$ is not a finite measure space but a $\sigma$-finite one. We then find a sequence $(X_n)_n$ of measurable subsets $X_n \subset X$ such that $\mu(X_n) < \infty$ and $\bigcup_{n \in \mathbb{N}} X_n = X$. Fix an arbitrary $n \in \mathbb{N}$. As $X_n$ is a subset of $X$, we conclude that $f_n \rightharpoonup g$ in $L^q(X_n, \mu)$ and $f_n \to f$ almost everywhere in $X_n$, and hence, by Step 1, we deduce that $f = g$ almost everywhere in $X_n$. As $n $ was arbitrary, we find that $f = g$ a.e. in $\bigcup_{n \in \mathbb{N}} X_n = X$. This proves the claim.