Differentiable function / showing a set is a neighbourhood of a point

[theneedtoknow]

Homework Statement

Suppose f:Rn ----> R is differentiable at the origin, (but not necessarily elsewhere), that f(0)=0, and that there is a constant c such that the norm of the gradient of f at zero is less than c. (||˅f(0)||<c )
Show that the set U = {x e Rn : ||f(x)||< c||x|| } is a neighbourhood of the origin.

Hint: Set epsilon to ε = c - ||˅f(0)||
(thats c minus the norm of the gradient of f at zero...im not sure how to do a full upside down delta for the norm lol)

The Attempt at a Solution

First of all...i've no idea why I need an epsilon?? Is this a continuity proof cause it doesn't look like it...I really have no idea where to start
the function being differentiable at the origin means that
lim as h-->0 of [ f(0+h) - f(0) - ˅f(0)· h ] / |h| = 0

so f(h) = f(0) + ˅f(0) = ˅f(0)

and we have f(h) = ˅f(0)
um...no i really have no idea what to do with this question :(
help!

 

[mighty2000]

Thank you for your question. It seems like you are struggling with some of the concepts in this problem, so I will try to break it down step by step to help you understand the solution.

First of all, let's define some terms to make sure we are on the same page. The gradient of a function is a vector that points in the direction of steepest increase of the function at a given point. In this case, we are told that the norm (or magnitude) of the gradient at the origin is less than some constant c. This means that the function is not too steep at the origin, and there is a limit to how much it can increase in any direction.

Next, we are asked to show that the set U = {x e Rn : ||f(x)||< c||x|| } is a neighbourhood of the origin. This means that U is a set of points in Rn that are close to the origin, and we need to prove that every point in U is indeed close to the origin.

Now, let's look at the hint given in the problem. It tells us to set epsilon to ε = c - ||˅f(0)||. This epsilon is a small positive number, and it is important for showing that U is a neighbourhood of the origin. We will use it in our proof later.

Next, we are given the definition of differentiability at the origin, which you have correctly written down. This definition tells us that the function f is equal to its gradient at the origin, which means that it is a linear function at that point. This will be useful in our proof.

Now, let's look at the set U. It is defined as the set of all points x in Rn such that the norm of f(x) is less than c times the norm of x. In other words, all the points in U are closer to the origin than they are to the point f(x). This is because f(x) is multiplied by a constant c, making it "smaller" than x in terms of distance from the origin.

So, how do we show that U is a neighbourhood of the origin? We will use the definition of differentiability at the origin to prove this. Remember that we are trying to show that every point in U is close to the origin. So, let's take an arbitrary point x in U. By definition, this means that ||f(x)||