# $\lim\inf \int\limits_{-y}^{y} |f(x,y)| dx =0.$

Let $$T$$ be the triangle $$\{(x,y): 0\le |x| \le y\le 1\},$$ and $$\mu$$ be the restriction of planar Lebesgue measure to $$T$$. Suppose that $$f\in L^2(T, \mu).$$ Prove that $$\lim\inf\limits_{y\to 0^+} \int\limits_{-y}^{y} |f(x,y)| dx=0.$$

I have tried to use Cauchy-Schwarz inequality to obtain $$\int\limits_{-y}^{y} |f(x,y)| dx\le \sqrt{2y}\left( \int\limits_{-y}^{y} |f(x,y)|^2dx\right )^{1/2}.$$

Since $$f$$ is in $$L^2(T)$$ the above integral is finite for almost every $$y$$. But, that is not enough to argue that it goes to $$0$$ as $$y\to 0$$.

Any hint would be appreciated.

## 1 Answer

Prove by contradcition. Suppose the result is not true. Then there exists $$a,b >0$$ such that $$\int_{-y}^{y} |f(x,y)| dx >a$$ for $$0< y. This gives $$a^{2} < 2y \int_{-y}^{y} |f(x,y)|^{2} dx$$. Let $$0 and integrate w.r.t. $$y$$ from $$0$$ to $$r$$. This gives $$ra^{2} \leq 2r \int _0^{r}\int_{-y}^{y} |f(x,y)|^{2} dxdy$$. This leads to a contradiction when you let $$r \to 0$$. [If $$g$$ is integrable and $$\mu (E) \to 0$$ then $$\int_{E} g d\mu \to 0$$].

• Thanks! It’s very neat, I was thinking along the similar lines but couldn’t actually get to the correct idea. – WhoKnowsWho Aug 14 at 14:21