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.


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 $0< y<b$. This gives $a^{2} < 2y \int_{-y}^{y} |f(x,y)|^{2} dx$. Let $0<r<b$ 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 $\int_{E} g d\mu \to 0$].

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

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