separately continuous functions $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ but nowhere continuous

Is there a functions $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ separately continuous but nowhere continuous?

There is a general question Here and this is unanswered.

Solution 1:

The research interests of Baire (1874-1932) in continuity of functions (and the corresponding 'Baire classes') have started with the question on the relationship between continuity of a function of two variables in each argument separately and its joint continuity in the two variables. Baire have shown that such function are of first Baire class. This was generalized by Lebesgue in the following sense: Any separately continuous function in $n$ variables is of Baire class $n-1$.

Let $f \colon \mathbb{R}^2 \rightarrow \mathbb{R}$ be any separately continuous function. Then $f$ cannot be nowhere continuous. In fact, the set of continuity is of first class and thus dense.

Proof: We prove that $f$ is the limes of continuous functions. For this let $$g(x) = \begin{cases} 1-|x| & \text{ if } |x| \ \leq 1 \\ 0 & \text{ otherwise} \end{cases}$$ Moreover define following partion of unity: $$g_n(x) := \frac{g(x-n)}{\sum_{m \in \mathbb{Z}} g(x+m)}$$ Note that $g_n(n) =1$ and $g_n$ is supported on $[n-1,n+1]$ and $\sum_{n \in \mathbb{Z}}g(x+n) =1$. We define $$f_n(x,y) := \sum_{k=-\infty}^\infty f(x,k \cdot 2^{-n}) g_k(2^ny).$$ Then $f_n$ is a continuous function and $f_n \rightarrow f$ pointwise, i.e. $f$ is of first Baire category (by definition pointwise limes of continuous function). It is well-known that in this case the set of points of discontinuity is of the first category. Because $\mathbb{R}^2$ is a complete space, and thus a Baire space, the set of continuity is dense.

We can construct an examples, which is (for example) not continuous on $\mathbb{Q} \times \mathbb{Q}$, by using a function which is separately continuous but not continuous in one point. One frequently used example is $$f(x,y) := \begin{cases} \frac{xy}{x^2+y^2} & \text{ if } (x,y) \ne (0,0) \\ 0 & \text{ if } (x,y) = (0,0) \end{cases}.$$ Note that because of $|xy| \le x^2 +y^2$, we always have $|f(x,y)| \le 1$. Let $(q_n)_{n \in \mathbb{N}}$ be an enumeration of $\mathbb{Q}$. Define $$g(x,y) = \sum_{n,k=1}^\infty \frac{1}{2^{n+k}} f(x-q_n,y-q_n).$$ This function is separately continuous, but not continuous in any point of $\mathbb{Q} \times \mathbb{Q}$.