Question

Show that the continuity of \(\mathbf{f}^{\prime}\) at the point a is needed in the inverse function theorem, even in the case \(n=1:\) If $$ f(t)=t+2 t^{2} \sin \left(\frac{1}{t}\right) $$ for \(t \neq 0\), and \(f(0)=0\), then \(f^{\prime}(0)=1, f^{\prime}\) is bounded in \((-1,1)\), but \(f\) is not on ny-neighborhood of 0

Short answer

Despite being increasing near 0 and having a finite slope in the interval (-1,1), the function f(t) = t + 2t^2sin(1/t) is not one-to-one in any neighborhood of 0. This is due to the oscillatory behavior of the function caused by the sin(1/t) term near 0. As shown through the sequences of points t_k and t'_k, we can find two distinct points t_1 and t_2 close to 0 such that f(t_1) = f(t_2). This shows the necessity of continuity of f'(t) at a=0 even for n=1 in the inverse function theorem.

Step-by-step solution

Find the derivative of the function \(f(t)\)

To find the derivative of the function \(f(t)\), we will use the sum and product rules of differentiation. The given function is: $$ f(t)=t+2t^2\sin\left(\frac{1}{t}\right) $$ The derivative of the function with respect to \(t\) is: $$ f'(t)=1+4t\sin\left(\frac{1}{t}\right)-2t\cos\left(\frac{1}{t}\right), $$ where the first term comes from differentiating \(t\) and the second and third terms come from applying the product rule to \(2t^2\sin\left(\frac{1}{t}\right)\).

Analyze the behavior of \(f(t)\) in some neighborhood of 0

It is given that \(f'(0)=1\) and \(f'(t)\) is bounded in \((-1,1)\). This means that the function is increasing near 0, and it has a finite slope in the interval \((-1,1)\). However, the function is oscillatory near \(0\) due to the presence of the \(\sin\left(\frac{1}{t}\right)\) term, which changes signs rapidly near zero.

Show that \(f(t)\) is not one-to-one in any neighborhood of 0

To show that \(f(t)\) is not one-to-one in any neighborhood of 0, we will show that there exist two distinct points \(t_1\) and \(t_2\) close to 0 such that \(f(t_1)=f(t_2)\). Consider the sequence of points \(t_k=\frac{1}{2k\pi}\) for \(k=1,2,3,...\). Since \(\sin\left(\frac{1}{t_k}\right)=\sin(2k\pi)=0\), we have $$ f(t_k)=t_k+2t_k^2\sin\left(\frac{1}{t_k}\right)=t_k. $$ Now, consider the sequence of points \(t'_k=\frac{1}{(2k+1)\pi}\) for \(k=1,2,3,...\). Since \(\sin\left(\frac{1}{t'_k}\right)=\sin((2k+1)\pi)=0\), we have $$ f(t'_k)=t'_k+2t'_k^2\sin\left(\frac{1}{t'_k}\right)=t'_k. $$ Therefore, as we can have both sequences with the same function values as \(k\) gets larger and \(\lim_{k\to\infty}t_k=\lim_{k\to\infty}t'_k=0\), we can find two distinct points \(t_1\) and \(t_2\) close to 0 such that \(f(t_1)=f(t_2)\). Hence, \(f(t)\) is not one-to-one in any neighborhood of 0, which shows the necessity of the continuity of \(f'(t)\) at \(a=0\) in the inverse function theorem even for \(n=1\).

Key concepts

Continuity of Derivatives

In mathematics, the concept of continuity of derivatives is crucial in various areas, including the inverse function theorem. When we talk about the continuity of a derivative at a particular point, it implies that the derivative function approaches a specific value as we get closer and closer to that point. This condition is necessary to apply the inverse function theorem.
In the problem given, the derivative of the function, denoted as \( f'(t) \), is bounded in the interval \((-1,1)\). It may suggest regular behavior, but due to oscillation, it lacks the necessary continuity near zero.
This lack of continuity means that even though the derivative exists at \( t=0 \), it doesn't behave in an orderly manner as it nears zero. Therefore, the function \( f(t) \) can't be locally inverted around the origin. This is precisely why continuity of the derivative is imperative in the inverse function theorem, to ensure that the function remains smooth and predictable for precise local inversion.

One-to-One Function

A one-to-one function is a type of function where each input corresponds to a unique output. In other words, no two different input values can produce the same output value. This characteristic is crucial in many proofs and applications, especially concerning the inverse of a function.
In the context of this exercise, we're asked to demonstrate that the function \( f(t) \) is not one-to-one near the origin. While analyzing the behavior of \( f(t) \), it becomes clear that due to the involvement of the oscillating factor \( \sin\left(\frac{1}{t}\right) \), the function fails to reserve one-to-one correspondence around zero.
We illustrate this by considering two sequences of points \( t_k \) and \( t'_k \) approaching zero. These points repeatedly map to the same function value, showcasing multiple inputs leading to the same output. Therefore, \( f(t) \) defies the one-to-one property in its neighborhood around zero, highlighting the necessity for continuity as it influences local invertibility.

Differentiation Rules

Differentiation rules are fundamental tools in calculus used to find the derivatives of functions. They include various techniques such as the product rule, quotient rule, and chain rule. These rules simplify the process of differentiating complex functions comprised of basic differentiable components.
In the exercise provided, we used the sum and product rules to differentiate the function \( f(t) = t + 2t^2\sin\left(\frac{1}{t}\right) \). The sum rule allows us to differentiate the sum of two functions separately, while the product rule applies when handling the product of two functions.
For example, when differentiating \( 2t^2\sin\left(\frac{1}{t}\right) \), we use the product rule since we have a product of \( 2t^2 \) and \( \sin\left(\frac{1}{t}\right) \). These differentiation strategies are critical in determining \( f'(t) = 1 + 4t\sin\left(\frac{1}{t}\right) - 2t\cos\left(\frac{1}{t}\right) \), helping us analyze how the function behaves, especially around \( t=0 \).

Oscillatory Functions

Oscillatory functions exhibit regular variations, such as periodic fluctuations, over a range of values. These types of functions are characterized by repeating patterns or cycles. Examples include trigonometric functions such as sine and cosine, which swing back and forth in a regular interval.
In the context of the given function \( f(t) = t + 2t^2\sin\left(\frac{1}{t}\right) \), the term \( \sin\left(\frac{1}{t}\right) \) introduces an oscillatory behavior into the function as \( t \) approaches zero. Such oscillations can be very rapid and erratic near zero, resulting in a function that behaves unpredictably in that area.
This oscillatory nature is responsible for preventing \( f(t) \) from being one-to-one near zero. The rapid sign changes of \( \sin\left(\frac{1}{t}\right) \) make it impossible for the function to maintain a single direction or monotonicity. Consequently, this blocks the possibility of a smooth inversion and highlights the necessity of specific continuity criteria for derivatives in theoretical applications like the inverse function theorem.