Calculus Problems

Question

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\\2. Let $g$ be continuous on $[a, b]$ and $g(x) \in[a, b]$ for all $x \in[a, b] .$ Prove the following.

\\(i) $g$ has at least one fixed point in $[a, b]$

\\(ii) $\mathrm{If},$ in addition, $g^{\prime}(x)$ exists on $(\mathrm{a}, \mathrm{b})$ and a positive constant $\mathrm{k}<1$ exists

with $\left|g^{\prime}(x)\right| \leq k,$ for all $x \in(a, b),$ then prove that there is exactly one fixed

point in $[a, b]$

\\(iii) Use (ii) to show that $g(x)=\pi+0.5 \sin (x / 2)$ has a unique fixed point

on $[0,2 \pi]$

\\(iv) prove that the equation $x=\pi+0.5 \sin (x / 2)$ has exactly one real solution on $[0,2 \pi]$

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Answer 1

These theorems are well-known results;

for the first, we consider g(x) = f(x) - x

Since f(x) is in [a,b] on [a,b], f(a) >= a and f(b) <= b.

If either of these is an equality, then that is a fixed point, so you assume otherwise.

If f(a) > a, then g(a) = f(a) - a > 0

If f(b) < b then g(b) = f(b) - b < 0

If f is continuous on [a,b], so is g

As g(a) > 0, g(b) < 0, and g is continuous, g must take on all values between g(a) and g(b).

This includes 0. When g(x) = 0, f(x) = x, so we have a fixed point.

The uniqueness of the fixed point if the derivative exists and is less than one on the interval can be proven using the Mean-Value Theorem (in fact, the contrapositive).

Let there be a fixed point at y and z in [a,b]

Then, f(y) = y and f(z) = z

Then, the Mean-Value Theorem tell us that there must exist a c in [y,z] such that f'(c) = (f(z) - f(y))/(z-y) = (z-y)/(z-y) = 1.

Thus, there must be a point c in [y,z] where the derivative = 1. Yet, we are given that the derivative is less than one everywhere, which is a contradiction. Thus, there must be at most one fixed point (which we have proven exists), so there is a unique fixed point.

Then, the result that there is a unique fixed point on [0, 2 π] for f(x) =π + 1/2 sin(x/2) can be easily shown.

As sin ranges from -1 to 1, π + 1/2 sin(x/2) ranges from π - 1/2 to π + 1/2. As 0 < π - 1/2 and π + 1/2 < 2 π,

f(x) is bounded by [0, 2 π] on [0, 2 π], so there exists a fixed point.

f'(x) = 1/4 cos(x/2), which is <= 1/4 < 1 everywhere

Thus, the fixed point is unique.