i have the following question: Let $f$ be defined in the circle $x^2+y^2=1$, and for all $a$ in $(0,1)$:$$f(a,0)=f(0,a)=f(a,a)=a$$ Prove that $f$ is not differentiable at $(0,0)$. I've tried to suppose that $f$ is differentiable, so it follows that it's continuous at $(0,0)$, and then prove by definition that it can't be, but i get lost there. Any ideas?
Prove that f is not Differentiable
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continuity
proof-writing
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2Surely you mean $f$ is defined on the *disk* $x^2 + y^2 < 1$? – 2017-01-15
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0Your approach won't work since there are continuous functions that match the conditions. For example an inverted cone. Instead show that the partial derivatives at the origin cannot be continuous – 2017-01-15
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1@OpenBall it's ironic that you are the one to comment on this, given your name. I figured that must be what he's talking about, since the origin was mentioned. Also calculus 3 students don't often distinguish between the two, which can be confusing – 2017-01-15
1 Answers
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If it's differentiable at $0$, then it's continuous at $0$, so $\lim_{(a,b) \to (0,0)} f(a,b) = f(0,0)$. Since the limit exists, it should be the same along all paths, in particular along the path $b = 0$, hence $f(0,0) = \lim_{a \to 0} f(a,0) = \lim_{a\to 0} a = 0$. Now:
$$\frac{\partial f}{\partial x} (0,0) = \lim_{a \to 0} \frac{f(a+0,0) -f(0,0)}{a} = \lim_{a \to 0} 1 = 1$$
$$\frac{\partial f}{\partial y} (0,0) = \lim_{a \to 0} \frac{f(0,0+a) -f(0,0)}{a} = \lim_{a \to 0} 1 = 1$$
Recall that $f$ is differentiable at $0$ iff:
$$\lim_{(a,b) \to (0,0)} \frac{f(0+a,0+b) -f(0,0) - \frac{\partial f}{\partial x}(0,0) a - \frac{\partial f}{\partial y}(0,0) b}{\sqrt{a^2 + b^2}} = 0$$
In particular, the above limit should be $0$ along the path $b=a$. However,
$$\frac{f(0+a,0+b) -f(0,0) - \frac{\partial f}{\partial x}(0,0) a - \frac{\partial f}{\partial y}(0,0) b}{\sqrt{a^2 + b^2}} = \frac{f(a,b) - a - b}{\sqrt{a^2 + b^2}} =_{\text{ if }b = a} \frac{-a}{\sqrt 2 |a|} \not \to 0$$