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I am trying to figure out if the following function is in fact continuous: given $$f(x,y) = \left\{\begin{array}{cc} \frac{|y|-|x|}{y^2} & |x| < |y| \\ 0 & |x| \geq |y| \end{array}\right.$$ I am studying the continuity of the function $$g(y) = \int_{-1}^1 f(x,y) dx, y \in [-1,1].$$ $x$ and $y$ take on the domain $[-1,1].$ I am wondering what is the most reasonable way to begin exploring the continuity of this function, since it seems to me that the many absolute values that make up this particular function makes the splitting of the integral rather hectic. Any recommendations on how one goes about dealing with integrals of absolute value functions?

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  • $\begingroup$ Note: Let $a>0$. $|x| < a \iff -a < x < a$. Now set $a = |y|$. Since the integral is only over $x$, it's simple enough to see that it equals $\int_{-|y|}^{|y|} ...$. Now split this in the usual manner into integrals over which $x$ is positive and negative. $\endgroup$ – stochasticboy321 Jun 2 '16 at 0:42
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    $\begingroup$ The map $x \mapsto f(x,y)$ is even, hence you can write the integral as $2 \int_0^1$. Since $y \mapsto f(x,y)$ is even we see that $g$ is even, hence you only need to deal with $x,y \ge 0$ which should not be that 'hectic'. $\endgroup$ – copper.hat Jun 2 '16 at 0:43
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Regarding the the question about continuity, suppose you approach $0$ along the $y$ axis. $f(x,y)$ gets large as $y$ approaches $0.$ Now approach along the $x$ axis. $f(x,y) = 0$ the whole way in. $f(x,y)$ is not continuous.

For the integral. $f(x,y)$ is an even function (for both x and y). We can look at the 1st quadrant and quadruple the results.

$4\int_0^1\int_0^y \frac {y-x}{y^2} dx\, dy\\ 4\int_0^1 \frac {y-\frac12 x^2}{y^2}|_0^y dy\\ 4\int_0^1 \frac {y-\frac12 y^2}{y^2}dy\\ 4\int_0^1 \frac 1y - \frac12 dy\\ 4(\ln y - \frac 12 y)|_0^1$

which doesn't converge at $0$

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  • $\begingroup$ wolframalpha says that $\int_{-1}^{1} \frac{|y| - |x|}{y^2} dx = \frac{2|y| - 1}{y^2}$. Well $\lim_ {y \to 0} \frac{2|y| - 1}{y^2} = - \infty$. I believe I interpret the question different from you. $\endgroup$ – Olba12 Jun 2 '16 at 0:58
  • $\begingroup$ Oh I see my where I went wrong, sorry. :) $\endgroup$ – Olba12 Jun 2 '16 at 1:01

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