# If $f\in C^1, L^1$, does $\lim_{x\rightarrow \infty} f(x)=0$?

Problem: Prove or provide a counterexample: Let $m$ be the Lebesgue measure on $\mathbb{R}$. Suppose $f\in L^1(\mathbb{R}, m)$ is of class $C^1$, and that $f'\in L^1 (\mathbb{R}, m)$. Then $\lim_{x\rightarrow \infty} f(x) =0$.

Thoughts: Take the standard Gaussian function that integrates to 1. By changing the "variance" (using $e^{-x^2/c}$ for some $c$) we can vary the total integral while keeping the maximal value as 1. Translating these across the real line so that we have functions centered on each positive integer $n$ with total integral $\frac{1}{n^2}$ (and adding them together and using the monotone convergence theorem), we get a $C^\infty$ function in $L^1$. My scratchwork indicates the derivative is also in $L^1$, so this seems to provide a counterexample.

However, upon further review, my scratch work is incorrect...

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Check your scratchwork. – GEdgar Jul 2 '12 at 1:22

## 1 Answer

The theorem is true.

Let $g(x)=\int_x^{+\infty} |f'(t)| dt$, which is continuous and has limit 0 at $+\infty$.

Fix some $\varepsilon>0$ and pick $x_0$ such that $|g(x_0)|<\varepsilon/2$. Then if $|f(x)|\ge\varepsilon$ for some $x\ge x_0$, for all $y\ge x$ we have $$|f(y)-f(x)|=\left|\int_x^y f'(t) dt\right|\le \int_x^y |f'(t)| dt=g(y)-g(x)\le \varepsilon/2$$ so that $|f(y)|\ge \varepsilon/2$ for all $y\ge x$, which contradicts integrability of $f$.

So $|f(x)|<\varepsilon$ for all $x\ge x_0$ and $\lim_{x\to+\infty} f(x)=0$.

The reason your counterexample doesn't work is that the derivative increases when you shrink the bumps to keep $f$ integrable.

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