Continuous in every linear direction not implies continuous

Statement

For a function of two variables at a point

It is possible to have a function $f$ of two variables $x,y$ and a point $(x_0,y_0)$ in the domain of $f$ such that $f$ is continuous along every linear direction at $(x_0,y_0)$ (i.e., the function $h\mapsto (x_0+uh,y_0+vh)$ is continuous at $h=0$ for every vector $⟨u,v⟩$) but $f$ is not a continuous function.

Statement

For a function of two variables at a point

It is possible to have a function $f$ of two variables $x,y$ and a point $(x_0,y_0)$ in the domain of $f$ such that $f$ is continuous along every linear direction at every point but $f$ is not a continuous function.

Proof

Example

Consider the function:

$f(x,y):=\{\begin{array}{cc}0,& y\le 0\text{ or }y\ge x^2\\ (1-\frac{y}{x^2}),& 0<y<x^2\\ \end{array}$

Alternatively, we can describe it as:

$f(x,y):=\{\begin{array}{cc}max\{0,\frac{y}{x^2}(1-\frac{y}{x^2})\},& x\ne 0\\ 0,& x=0\\ \end{array}$

We first argue that $f$ is continuous at all points other than $(0,0)$:

• At any point with $y<0$, the function is the zero function around the point, hence is continuous.
• At any point with $y>x^2$, the function is the zero function around the point, hence is continuous.
• At any point with $0<y<x^2$, the function has the rational function description $\frac{y}{x^2}(1-\frac{y}{x^2})$ around the point, hence is continuous.
• At any point on the line $y=0$ other than the origin, there are two definitions of the function around the point: the definition 0 (from the $y\le 0$ side), and the definition $\frac{y}{x^2}(1-\frac{y}{x^2})$ (from the $0<y<x^2$ side). Both definitions evaluate to zero at the point, which is the same as the function value at the point.
• At any point on the curve $y=x^2$ other than the origin, there are two definitions of the function around the point: the definition 0 (from the $y\ge x^2$ side), and the definition $\frac{y}{x^2}(1-\frac{y}{x^2})$ (from the $0<y<x^2$ side). Both definitions evaluate to zero at the point, which is the same as the function value at the point.

We now argue that $f$ is continuous in every linear direction at $(0,0)$. It suffices to consider half-line directions because continuity from a linear direction is continuity from both half-line directions.

• For the half-line directions that are below or along the $y=0$ line, the function is identically zero along the half-line, so the limit at the origin is zero, which equals the zero.
• For the half-line directions that are above $y=0$, we note that, sufficiently close to the origin, this half-line is completely above the $y=x^2$ curve (explicitly, if the slope of the line is $m$, then the half-line is above $y=x^2$ for $\left|x\right|<\left|m\right|$). Thus, sufficiently close to the origin, $f$ looks like the zero function on this half-line. Thus, the limit at the origin is zero, which equals the value.

We finally demonstrate that $f$ is not continuous at $(0,0)$ by finding a curve approaching the origin along which the limit at the origin is not zero. Consider the curve:

$y=\frac{1}{2}{x}^2$

Consider the limit:

$\underset{x\to 0^{+}}{lim}f(x,y)=\underset{x\to 0^{+}}{lim}f(x,x^2/2)=\underset{x\to 0^{+}}{lim}(1/2)(1-1/2)=\underset{x\to 0^{+}}{lim}1/4=1/4\ne 0$

Intuitive explanation of example

Intuitively, this example function is zero on a very large subset of the domain, and the set of points where it is nonzero is a narrow squished subset of the plane that, near the origin, is too small to be detected by straight lines.