Topological Spaces and Continuous Functions

Question

I have no idea where to even start with this problem. We were going over metric spaces in class and this never showed up in the chapter. If someone could help me out it would be much appreciated!

Let $f:[0,\infty) \to [0,\infty)$ be a continuously twice-differentiable, strictly increasing, and concave (also called concave down; i.e. $f''< 0)$ function such that $f(0) = 0$. $\\$

A: Show that the following function $ϕ: [0, ∞) → R$ is decreasing for any fixed $t > 0$:

$ϕ(x) = { f(x + t) − f(x)\over t }$

B: Prove for $x ≥ 0$ and $t > 0$ that:

$f(x + t) ≤ f(x) + f(t)$.

C: Show that the rational function ${x\over 1+x}$ satisfies the inequality in part (b).

Answer 1

1. Check the derivative of $\phi$, and use the 2 conditions provided.

2. Use that $\phi$ is decreasing to conclude that $\phi(0) <= \phi(x)$. And rearrange to get the answer. Keep in mind that $t$ is positive so you can cancel that from the denominator.

3. Show that $f$ satisfies the conditions described by the exercise.

Answer 2

$\phi (x)=\frac {f(x+t)-f(x)} t =\frac 1 t\int_x^{x+t} f'(u)du=\int_0^{1} f'(x +tv)dv$ by the substitution $v=\frac {u-x} t$. Since $f'$ is decreasing it is obvious from this expression that $\phi$ is decreasing. For b) use the fact that $\phi (x) \leq \phi (0)$. c) is by direct verification.

Answer 3

(1).$\quad \phi'(x)=\frac {f'(x+t)-f'(x)}{t},$ which by the MVT is equal to $f''(x+s)$ for some $s\in (0,t).$ So $\phi'(x)<0.$

So if $y>0$ then $$\phi(x+y)-\phi(x)=y\cdot \frac {\phi(x+y)-\phi(y)}{y}$$ which by the MVT is equal to $y\phi'(x+z)$ for some $z\in (0,y).$ So $\phi(x+y)-\phi(x)<0.$

(2).$\quad$For a fixed $t>0$ let $g(x) =f(x+t)-f(x)-f(t).$ Then $g(0)=-f(0)\le 0.$

By (1) we have $g'(y)=t\phi'(y)<0.$ For $x>0$ we have $$g(x)=g(0)+x\cdot\frac {g(x)-g(0)}{x}$$ which by the MVT is equal to $g(0)+xg'(y)$ for some $y\in (0,x).$ So $g(x)=g(0)+xg'(y)<g(0)\le 0.$

Answer 4

This is for part A.

Show that is decreasing.

is nothing more than the subtraction of two functions.

The function is just a function being shifted to the left t units. Because t is always positive the shift will always be a horizontal shift to the left graphically.

The function is the same function only without the horizontally shift.

Because essentially these are the same functions the one with the shift has more time to grow or increase than the nonshifted one. So we can conclude that:

The shifted function is always greater than the non-shifted function. In other words:

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Thus we can conclude that:

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Which means that between these two functions there must be some vertical distance between them that exists for any given particular x value.

Now we have three cases. These two functions can either diverge away from each other, converge on each other, or have a constant distance between them.

Consider the convergent case. If the two functions are converging then the distance values between them keep geting smaller and smaller.

Because this distance is getting smaller and smaller as x increases this is what allows us to conclude that is actually decreasing.