- Let \(f\) be a real-valued function defined on \(\mathbb{R}^n\). Define differentiability of \(f\) at a point \(p\).
- Show that the function \(f: \mathbb{R}^2 \to \mathbb{R}\) defined by \[f(x,y) = \frac{xy}{x^2+y^2} \text{ for } (x,y) \neq (0,0), \text{ and } f(0,0) = 0\] is not differentiable at \((0,0)\).
Solution
1. Part (a): Definition of differentiability:
1.1. A function \(f: \mathbb{R}^n \to \mathbb{R}\) is differentiable at a point \(p \in \mathbb{R}^n\) if there exists a linear transformation \(L: \mathbb{R}^n \to \mathbb{R}\) (or equivalently a gradient vector \(\nabla f(p) \in \mathbb{R}^n\) such that \(L(h) = \langle \nabla f(p), h \rangle\)) satisfying:\[ \lim_{h \to 0} \frac{|f(p + h) - f(p) - L(h)|}{\|h\|} = 0. \]Equivalently, \(f(p + h) = f(p) + L(h) + o(\|h\|)\) as \(\|h\| \to 0\).
standard definition of Fréchet differentiability on \(\mathbb{R}^n\).
2. Part (b): Non-differentiability of \(f(x, y)\) at \((0, 0)\):
2.1. If a function \(f\) is differentiable at \(p\), then \(f\) is continuous at \(p\).
\(|f(p+h) - f(p)| \le |L(h)| + o(\|h\|) \le \|L\| \|h\| + o(\|h\|) \to 0\) as \(h \to 0\).
2.2. We evaluate the limit of \(f(x, y) = \frac{xy}{x^2 + y^2}\) as \((x, y) \to (0, 0)\) along lines \(y = mx\):For \(x \neq 0\):\[ f(x, mx) = \frac{x(mx)}{x^2 + (mx)^2} = \frac{m x^2}{x^2(1 + m^2)} = \frac{m}{1 + m^2}. \]
algebraic substitution \(y = mx\).
2.3. Along the line \(y = 0\) (\(m = 0\)), \(\lim_{x \to 0} f(x, 0) = 0\).Along the line \(y = x\) (\(m = 1\)), \(\lim_{x \to 0} f(x, x) = \frac{1}{1 + 1^2} = \frac{1}{2} \neq f(0, 0)\).Since the limit depends on the path of approach, \(\lim_{(x, y) \to (0, 0)} f(x, y)\) does not exist.Thus \(f\) is discontinuous at \((0, 0)\).
non-uniqueness of directional limits.
2.4. Since \(f\) is not continuous at \((0, 0)\), by 1 \(f\) cannot be differentiable at \((0, 0)\).
contrapositive of differentiability implies continuity.
3. Conclusion:\(f\) is defined to be differentiable if its linear approximation error is \(o(\|h\|)\), and \(f(x, y) = \frac{xy}{x^2+y^2}\) is not differentiable at \((0, 0)\) due to discontinuity. Q.E.D.
1 and 2.