Prove that $\lim_{n\to \infty}\|f+g_n\|_1 = \|f\|_1+\|g\|_1$

Question

Let $f,g,g_n\in L^1(\mathbb{R})$, where \begin{align} g_n(x) = g(x-n)\text{ for all }x \in \mathbb{R}. \end{align} Prove that $$\lim_{n\to \infty}\left|f+g_n\right|_1 = \left|f\right|_1 + \left|g\right|_1 .$$

It is probably simple but not clicking me right now, appreciate a hint.

Answer 1

Recall that the space of continuous compactly supported functions $C_{00}$ is $L^{1}$ dense. Now pick $\varphi,\psi\in C_{00}$ such that $\|f-\varphi\|_{L^{1}}$ and $\|g-\psi\|_{L^{1}}$ are small.

We see that \begin{align*} &\big|\|f+g_{n}\|_{L^{1}}-(\|f\|_{L^{1}}+\|g\|_{L^{1}})\big|\\ &\leq\big|\|f+g_{n}\|_{L^{1}}-\|\varphi+\psi_{n}\|_{L^{1}}\big|+\big|\|\varphi+\psi_{n}\|_{L^{1}}-(\|\varphi\|_{L^{1}}+\|\psi\|_{L^{1}})\big|\\ &~~~~~~~~+\big|\|\varphi\|_{L^{1}}+\|\psi\|_{L^{1}}-(\|f\|_{L^{1}}+\|g\|_{L^{1}})\big|, \end{align*} and \begin{align*} \big|\|f+g_{n}\|_{L^{1}}-\|\varphi+\psi_{n}\|_{L^{1}}\big|\leq\|f-\varphi\|_{L^{1}}+\|g_{n}-\psi_{n}\|_{L^{1}}, \end{align*} and \begin{align*} \|g_{n}-\psi_{n}\|_{L^{1}}=\int_{\mathbb{R}}|g(x-n)-\psi(x-n)|dx=\int_{\mathbb{R}}|g(x)-\psi(x)|dx=\|g-\psi\|_{L^{1}}, \end{align*} so the term $\big|\|f+g_{n}\|_{L^{1}}-\|\varphi+\psi_{n}\|_{L^{1}}\big|$ is small, whereas \begin{align*} \big|\|\varphi\|_{L^{1}}+\|\psi\|_{L^{1}}-(\|f\|_{L^{1}}+\|g\|_{L^{1}})\big|\leq\|f-\varphi\|_{L^{1}}+\|g-\psi\|_{L^{1}}, \end{align*} which is also small.

So we need only to concentrate on $\big|\|\varphi+\psi_{n}\|_{L^{1}}-(\|\varphi\|_{L^{1}}+\|\psi\|_{L^{1}})\big|$. In other words, \begin{align*} \lim_{n\rightarrow\infty}\|\varphi+\psi_{n}\|_{L^{1}}=\|\varphi\|_{L^{1}}+\|\psi\|_{L^{1}} \end{align*} is what we are to show for.

Now let $M>0$ such that $\text{supp}(\varphi),\text{supp}(\psi)\subseteq\{|x|\leq M\}$, then for $n\geq 3M$, we have \begin{align*} \|\varphi+\psi_{n}\|_{L^{1}}&=\int_{\mathbb{R}}|\varphi(x)+\psi(x-n)|dx\\ &=\int_{|x|\leq M}|\varphi(x)+\psi(x-n)|dx+\int_{|x|>M}|\varphi(x)+\psi(x-n)|dx, \end{align*} while for $|x|\leq M$, we have $|x-n|\geq n-|x|\geq 3M-M=2M$, so $\psi(x-n)$ vanishes for all such $x$, this yields that \begin{align*} \int_{|x|\leq M}|\varphi(x)+\psi(x-n)|dx=\int_{|x|\leq M}|\varphi(x)|dx=\|\varphi\|_{L^{1}}. \end{align*} On the other hand, \begin{align*} \int_{|x|>M}|\varphi(x)+\psi(x-n)|dx&=\int_{|x|>M}|\psi(x-n)|dx\\ &=\int_{\mathbb{R}}|\psi(x-n)|dx-\int_{|x|\leq M}|\psi(x-n)|dx\\ &=\int_{\mathbb{R}}|\psi(x)|dx-\int_{|x|\leq M}|\psi(x-n)|dx\\ &=\|\psi\|_{L^{1}}-\int_{|x|\leq M}|\psi(x-n)|dx, \end{align*} once again we have $|x-n|\geq 2M$, so \begin{align*} \int_{|x|>M}|\varphi(x)+\psi(x-n)|dx=\|\psi\|_{L^{1}}, \end{align*} this shows that $\|\varphi+\psi_{n}\|_{L^{1}}$ is eventually $\|\varphi\|_{L^{1}}+\|\psi\|_{L^{1}}$, we are done.

Answer 2

By the triangle inequality and translation invariance, we have $$\|f+g_n\| \leq \|f\| + \|g_n\| = \|f\|+ \|g\|,$$ which proves one inequality. For the other, suppose $f, g \geq 0$. Note $$f+g_{2n} \geq f \cdot 1_{(-\infty,n)} + g_{2n} \cdot 1_{(n, \infty)}= f \cdot 1_{(-\infty,n)} + g \cdot 1_{(-n, \infty)}.$$ Taking norms and sending $n \to \infty$ (with the dominated convergence theorem) gives the other inequality. Try to do the general case yourself.

Edit: The intuition for why the statement in the problem is correct as follows. Integrable functions behave like the "bell curve" $e^{-x^2}$ in the sense that most of their mass is "in the middle of $\mathbb{R}$," i.e. for any integrable function $f$, we have $$\lim_{R \to \infty} \int_{|x| \geq R} f = 0.$$ Thus, if $f$ and $g$ are integrable and we translate $g$ by a large number $n$, then the "area" under $f+g$ should roughly be that under $f$ plus that under $g$.