Problem Consider the (real-valued) function f: R² R defined by 0 for (x, y) = (0,0), for (x, y) (0,0). f(x, y) = 23 x² +

[answerhappygod]

Problem Consider The Real Valued Function F R R Defined By 0 For X Y 0 0 For X Y 0 0 F X Y 23 X 1 (105.22 KiB) Viewed 17 times

Problem Consider the (real-valued) function f: R² R defined by 0 for (x, y) = (0,0), for (x, y) (0,0). f(x, y) = 23 x² + y² (a) Prove that the partial derivatives Dif (Actually, f is continuous! Why?) af af = and D₂f = are bounded in R². ər ду (b) Let v = (v₁, v₂) E R2 be a unit vector. By using the limit-definition (of directional derivative), show that the directional derivative (Def)(0,0) = (Df)((0,0), v) exists (as a function of v), and that its absolute value is at most 1. [Actually, by using the same argument one can (easily) show that f is Gâteaux differentiable at the origin (0,0).] (c) Lety: R→ R2 be a differentiable function [that is, y is a differentiable curve in the plane R2] which is such that y(0) (0,0), and y'(t) (0,0) whenever y(t) = (0,0) for some t E R. Now, set g(t) = f(y(t)) (the composition of f and y), and prove that (this real- valued function of one real variable) g is differentiable at every tER. Also prove that if y E C¹ (R, R2), then g E C¹ (R, R). [Note that this shows that f has "some sort of derivative" (i.e., some rate of change) at the origin whenever it is restricted to a smooth curve that goes through the origin (0,0).] (d) In spite of all this, prove that f is not (Fréchet) differentiable at the origin (0,0). (Hint: Show that the formula (D,f) (0,0) = ((Vf)(0, 0), v) fails for some direc- tion(s) v. Here (,) denotes the standard dot product in the plane R².) [Thus, f is not (Fréchet) differentiable at the origin (0,0). For, if f were differentiable at the origin, then the differential f'(0, 0) would be completely determined by the partial derivatives of f; i.e., by the gradient vector (Vf)(0,0). Moreover, one would have that (D,f) (0,0) = ((Vf)(0,0), v) for every direction v; as discussed in class!]