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I'm struggling a bit with Chapter 10 of Rudin's "Principles of Mathematical Analysis," and I was hoping to get some help here.

I'll post the problem and my current progress.

Exercise 2: For $i = 1, 2, \ldots$ let $\phi_i \in \mathcal{C}(\mathbb{R}^1)$ have support in $(2^{-i}, 2^{1-i})$ such that $\int \phi_i = 1$. Let $f(x, y) = \displaystyle\sum_{i=1}^\infty [\phi_i (x) - \phi_{i+1} (x)] \phi_i (y)$. Then $f$ has compact support in $\mathbb{R}^2$, $f$ is continuous except at $(0, 0)$, and $\int dy \int f(x, y) dx = 0$ but $\int dx \int f(x, y) dy = 1$. Observe that $f$ is unbounded in every neighborhood of $(0, 0)$.

Progress

First, note that the supports of the $\phi_i$ are disjoint. Suppose $\phi_i (x) \neq 0$ on the set $S_i$ for each $i$. Then if $x \not\in \displaystyle\cup_{i = 1}^\infty S_i$, or if $y \not\in \displaystyle\cup_{i = 1}^\infty S_i$ we have $f(x, y) = 0$. If $x \in S_i$ for some $i$ then $x \not\in S_j$ for $j \neq i$, so $f(x, y) = -\phi_i (x) \phi_{i-1} (y) + \phi_i (x) \phi_{i} (y) = \phi_i(x) [\phi_{i} (y) - \phi_{i-1} (y)]$.

Note that $f(0, 0) = 0$ since $0$ is not in the support of any $\phi_i$. Then if we can show that $f$ is unbounded in every neighborhood of $(0, 0)$, by the epsilon-delta definition of continuity we know that $f$ is not continuous at $(0, 0)$.

Since $\int \phi_i = 1$ we know that the support for each $\phi_i$ must be non-empty.

Examine the neighborhood of radius $r$ centered at $(0, 0)$. If we can show that $f$ is unbounded on any neighborhood of radius $r < 1$ centered at $(0, 0)$, then by extension it is unbounded on any neighborhood of radius $r \ge 1$ centered at $(0, 0)$. Then suppose $2^{-a} \le r < 2^{1-a}$.

Suppose for sake of contradiction that $f$ is bounded in this neighborhood. Then let $\alpha, \beta$ be the values in $\mathbb{R}^1$ such that $f(\alpha, \beta) \ge f(x, y)$ for all $x, y \in \mathbb{R}^1$ with $|(x, y) - (0, 0)| < r$.

Suppose $\alpha \in S_i$. We must then have $\beta \in S_{i-1}$ or $\beta \in S_{i}$. If $\beta \in S_{i-1}$ then $f(\alpha, \beta) = -\phi_i (\alpha) \phi_{i-1} (\beta)$, and if $\beta \in S_i$ then $f(\alpha, \beta) = \phi_i (\alpha) \phi_i (\beta)$.

(~~~ I don't know how to proceed from here. I need to generate an ordered pair in $\mathbb{R}^2$ that creates a larger value for $f$, but I don't know how to judge the magnitude of $\phi_i (\alpha), \phi_i (\beta)$, or $\phi_{i-1} (\beta)$. I think I need to use $\int \phi_i = 1$, but to be totally honest I don't know what that means, or how to use it. ~~~)

Since the $\phi_i$ are continuous, we know that the $S_i$ are collections of intervals. Then $(0, 1) \backslash \cup_{i = 1}^\infty S_i$ is also a collection of intervals.

We showed earlier that $f(x, y) \neq 0$ iff for some $i$ we have $x \in S_i$ and $y \in S_{i} \cup S_{i - 1}$. Then $f(x, y) \neq 0$ over a set whose closure is a collection of $2$-cells, implying by definition that the support of $f(x, y)$ is a collection of $2$-cells. Since $2$-cells are compact and the union of a collection of compact sets is compact, the support of $f(x, y)$ is compact.

To show that $f$ is continuous (other than at the origin), we can examine $f$'s behavior in cases. If $x > 1$ or $x < 0$ and $y > 1$ or $y < 0$, then we can construct a disc in $\mathbb{R}^2$ centered at $(x, y)$ that does not intersect the set determined by $0 < x, y < 1$, so $f(x, y) = 0$ on this neighborhood since $\cup_{i = 1}^\infty S_i \subseteq (0, 1)$; this satisfies the epsilon-delta definition of continuity.

If $x = 1$ or $y = 1$ then $f(x, y) = 0$. Since $S_1$ is a collection of intervals in $\mathbb{R}^1$, it has a maximum; let it be $s$. Since $\overline{S_1} \in (1/2, 1)$, and since $\overline{S_1}$ is a collection of closed intervals, we know that $s < 1$. Then if $x = 1$ or $y = 1$, we have that $f(x, y) = 0$ on the neighborhood centered at $(x, y)$ with radius $1-s$. This satisfies the epsilon-delta definition of continuity.

Else if $x = 0$ and $y \neq 0$ then $f(x, y) = 0$. Suppose $y \in S_i$ for some $i$; then it is in some interval $[a, b] \in S_i$. Then consider the region defined by $(x', y') : x' < 2^{-(i+1)}, y' \in [a, b]$. We can inscribe a disk in this region centered at $(x, y)$, and we have $f(x', y') = 0$ over this region. This satisfies the epsilon-delta definition of continuity.

On the other hand, suppose $y \not\in S_i$ for any $i$; then it is in some interval $[a, b]$ such that $\phi_i (y) = 0$ for all $y \in [a, b]$. Through a similar construction we can create a disk over which $f$ takes the value of zero alone. This satisfies the epsilon-delta definition of continuity.

Else if $y = 0$ and $x \neq 0$ then $f(x, y) = 0$. Suppose $x \in S_i$ for some $i$; then it is in some interval $[a, b] \in S_i$. Then consider the region defined by $(x', y') : x' \in [a, b], y' < 2^{-i}$. We can inscribe a disk in this region centered at $(x, y)$, and we have $f(x', y') = 0$ over this region. This satisfies the epsilon-delta definition of continuity.

On the other hand, suppose $x \not\in S_i$ for any $i$; then it is in some interval $[a, b]$ such that $\phi_i (x) = 0$ for all $x \in [a, b]$. Through a similar construction we can create a disk over which $f$ takes the value of zero alone. This satisfies the epsilon-delta definition of continuity.

This shows the continuity of $f$ on all points other than $(0, 0)$.

(~~~ I don't know what the notation $\int dx \int f(x, y) dy$ means... ~~~)

Thanks so, so, so, so much for your help!

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1 Answer 1

$Hint:$ Since $supp(\phi_i)\subseteq (2^{-i},2^{1-i})$ then $\int_{2^{-i}}^{2^{1-i}} \phi_i=1$. On the other hand, $\max(\phi_i)\cdot 2^{-i}\geq \int_{2^{-i}}^{2^{1-i}} \phi_i$. Thus $\max(\phi_i)\geq 2^i$.

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