MHB Continuity of piecewise function of two variables

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The function f(x, y) is defined as 0 when y ≤ 0 or y ≥ x^4, and 1 when 0 < y < x^4. It is discontinuous at the point (0, 0) because, for any ε > 0, it is impossible to find a δ-neighborhood around (0, 0) where f remains within ε of its value at that point. Additionally, f is discontinuous along the curves y = 0 and y = x^4, as points on these curves yield different function values (0 or 1) within any neighborhood around them. The discussion emphasizes the challenge of demonstrating continuity due to the behavior of f near these critical points and curves. Overall, the function exhibits significant discontinuities in specified regions.
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The question looks like this.
Let $f(x, y)$ = 0 if $y\leq 0$ or $y\geq x^4$, and $f(x, y)$ = 1 if $0 < y < x^4 $.

(a) Show that $f$ is discontinuous at (0, 0)

(b) Show that $f$ is discontinuous on two entire curves.
In regarding (a), I know $f(x, y)$ is discontinuous on certain directions, but can't elaborate it in decent form.

In regarding (b), How can I show it?
 
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V150 said:
Let $f(x, y)$ = 0 if $y\leq 0$ or $y\geq x^4$, and $f(x, y)$ = 1 if $0 < y < x^4 $.

(a) Show that $f$ is discontinuous at (0, 0)
By definition, $f$ is continuous at $(0,0)$ if for every $\varepsilon>0$ there exists a $\delta>0$ such that if $(x,y)$ is located within distance $\delta$ from $(0,0)$, then $|f(x,y)-f(0,0)|=|f(x,y)|<\varepsilon$. Choose any $0<\varepsilon<1$. Can you find a circle around $(0,0)$ such that within that circle $f$ takes values $<\varepsilon$?

V150 said:
(b) Show that $f$ is discontinuous on two entire curves.
By the two curves, does the problem mean $y(x)=0$ and $y(x)=x^4$? Again, within every circle whose center lies on these curves, however small the radius is, there are points where $f$ returns 0 and other points where $f$ returns 1.
 

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