Exercise 10.2

Question

Exercise 2: For $i=1,2,3,\ldots$, let $\varphi_i\in\mathscr C(\R1)$ have support in $(2^{-i},2^{-i+1})$, such that $\int\varphi_i=1$. Put
    $$
        f(x,y)=\sum_{i=1}^\infty\big(\varphi_i(x)-\varphi_{i+1}(x)\big)\varphi_i(y)
    $$
    Then $f$ has compact support in $\R2$, $f$ is continuous except at $(0,0)$, and 
    $$
        \int dy\int f(x,y)\,dx=0\qquad\hbox{but}\qquad\int dx\int f(x,y)\,dy=1.
    $$
    Observe that $f$ is unbounded in every neighborhood of $(0,0)$.

analambanomenos · Accepted Answer

The $\varphi_i(x)\varphi_i(y)$ has support in the square $2^{-i}2^i$. Hence $f(x_i,x_i)=M_i^2>2^{i+1}$ diverges to $\infty$ as $i ightarrow\infty$, so $f$ is not continuous at $(0,0)$ and is unbounded in every neighborhood of $(0,0)$. We have \begin{align*} \int dy\int f(x,y)\,dx &= \sum_{i=1}^\infty\Bigg(\int\varphi_i(y)\,dy\Bigg)\Bigg(\int\varphi_i(x)\,dx-\int\varphi_{i+1}(x)\,dx\Bigg) \ &= \sum_{i=1}^\infty 1\cdot 0 = 0 \ \int dx\int f(x,y)\,dy &= \Bigg(\int\varphi_1(x)\,dx\Bigg)\Bigg(\int\varphi_1(y)\,dy\Bigg)\;+ \ &\qquad\sum_{i=2}^\infty\Bigg(\int\varphi_i(x)\,dx\Bigg)\Bigg(\int\varphi_i(y)\,dy-\int\varphi_{i-1}(y)\,dy\Bigg) \ &= 1\cdot 1 + \sum_{i=2}^\infty 1\cdot 0 = 1 \end{align*}

Exercise 10.2

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Exercise 10.2

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