# Thread: continous functions on intervals: existence of abs. min and max

1. ## continous functions on intervals: existence of abs. min and max

hello,

I have 2 problems here:

1) Let I = [0, pi/2] and let f:I->R be defined as f(x) = sup{x^2, cos x} for x E I. Show that their exists an absolute min point at u E I for f on I. Show that u is the solution to the equation cos x = x^2.

- attempt at a solution:
define f(x) = cos x when cos x>= x^2
= x^2 when x^2 > cos x
let x be such that f(x) = cos x. thus cos x is continuous for all x in R
let x be such that f(x) = x^2. x^2 is a polynomial and is continuous for all x in R
Does this imply that f(x) is continuous?
assuming that I proved that f(x) is continuous...
then since I is bounded and f is continuous on I, there exists an absolute min u in I for f on I.

i know that cos x = 1 @ 0 and x^2 = 0 at 0, so then f(0) = 1.
cos x decreases to 0, and x^2 increase to pi^2/4 (both monotonically to x = pi/2).
....

2) suppose f:R->R is continuous on R and that lim(x->infinity)f = 0 and lim(x-> - infinity) f = 0. Prove that f is bounded and that it attains either a minimum or a maximum.

- attempt at a solution:
since I = R is an interval, and f is continuous on I, then f(I) is also an interval.
show that f(I) is bounded.
NO idea

thanks

2. Originally Posted by sprinks13
hello,

I have 2 problems here:

1) Let I = [0, pi/2] and let f:I->R be defined as f(x) = sup{x^2, cos x} for x E I. Show that their exists an absolute min point at u E I for f on I. Show that u is the solution to the equation cos x = x^2.

- attempt at a solution:
define f(x) = cos x when cos x>= x^2
= x^2 when x^2 > cos x
let x be such that f(x) = cos x. thus cos x is continuous for all x in R
let x be such that f(x) = x^2. x^2 is a polynomial and is continuous for all x in R
Does this imply that f(x) is continuous?
assuming that I proved that f(x) is continuous...
then since I is bounded and f is continuous on I, there exists an absolute min u in I for f on I.

i know that cos x = 1 @ 0 and x^2 = 0 at 0, so then f(0) = 1.
cos x decreases to 0, and x^2 increase to pi^2/4 (both monotonically to x = pi/2).
....

2) suppose f:R->R is continuous on R and that lim(x->infinity)f = 0 and lim(x-> - infinity) f = 0. Prove that f is bounded and that it attains either a minimum or a maximum.

- attempt at a solution:
since I = R is an interval, and f is continuous on I, then f(I) is also an interval.
show that f(I) is bounded.
NO idea

thanks
1. Define $h(x)=\cos(x)-x^2$. It is apparent that $h(x)$ is continuous, and since $h(0)>0$ and $h\left(\frac{\pi}{3}\right)<0$ we can conclude by the Intermediate Value Theorem that there must be zero in $\left[0,\frac{\pi}{2}\right]~{\color{red}(*)}$.Note that $\cos(x)$ and $x^2$ are both monotonic on $\left[0,\frac{\pi}{2}\right]$ so there can be only one intersection, which we shall denote $\xi$. A quick sketch of the graph will then reveal that

$f(x)=\max\left\{\cos(x),x^2\right\}=\left\{ \begin{array}{rcl} \cos(x) & \mbox{if} & 0\leqslant x \leqslant \xi\\
x^2 & \mbox{if} & \xi

So the only possible point of discontinuity is $x=\xi$. To check this we first notice three things: $\xi$ is a limit point of $\left[0,\frac{\pi}{2}\right]$, that $\cos(x)$, and $x^2$ are left and right hand continuous respectively. So

$\lim_{x\to\xi^-}f(x)=\lim_{x\to\xi^-}\cos(x)=\cos(\xi)$

And

$\lim_{x\to\xi^+}f(x)=\lim_{x\to\xi^+}x^2=\xi^2$

And finally

$f(\xi)=\cos(\xi)$

Now remembering that $\xi$ is the unique point in $\left[0,\frac{\pi}{2}\right]$ such that $x^2=\cos(x)$ we can conclude that $\lim_{x\to\xi^-}f(x)=f(\xi)=\lim_{x\to\xi^+}f(x)$, thus $f(x)$ is continuous on $\left[0,\frac{\pi}{2}\right]$.

Now is here where I am not sure what level of math you are at. Tell me if this is beyond the scope of your course

Theorem: If $f:X\mapsto Y$ is continous and $X$ is compact, then so is $f(X)$

Proof: Let $\left\{\Psi_n\right\}$ be an open covering of $f(X)$. Using the fact that the inverse image of any open set in $Y$ must be open in $X$ we can see that $\left\{f^{-1}\left(\Psi_{n}\right)\right\}$ is an opening cover of $X$. But since $X$ is compact it follows that $X\subset f^{-1}\left(\Psi_{n_1}\right)\cup f^{-1}\left(\Psi_{n_2}\right)\cup\cdots$. Using the identity $f\left(f^{-1}\left(E\right)\right)\subset E$ we can concude that $f\left(X\right)\subset \Psi_{n_1}\cup\Psi_{n_2}\cup\cdots$ thus $f\left(X\right)$ is compact.

So knowing by the Heine-Borel Theorem that any closed set $[a,b]$ is compact we can conclude that $f\left([a,b]\right)$ is compact, thus closed and bounded, and due to its boundedness we know that the supremum and infimum are included.

So back to our actual problem since we have that $f:\left[0,\frac{\pi}{2}\right]\mapsto E$. So since $\left[0,\frac{\pi}{2}\right]$ is both bounded and closed it is compact, and since $f(x)$ is continuous it follows that so is $f\left(\left[0,\frac{\pi}{2}\right]\right)=E$, thus $\sup_{x\in\left[0,\frac{\pi}{2}\right]}f(x),\inf_{x\in\left[0,\frac{\pi}{2}\right]}f(x)\in E$, which is what was asked to be proven.

I'll come back later for number two.

Note: $\color{red}(*)$ was gotten by $f$ being continuous and $\left[0,\frac{\pi}{2}\right]$ being connected...in case you are in analysis.

3. Sorry, never talked about something being compact.

4. Originally Posted by sprinks13
Sorry, never talked about something being compact.
Ok, basically in $\mathbb{R}$ compact is synonomous with closed and bounded. So basically my theorem is saying that if $f$ is continuous then $f\left([a,b]\right)=E$ is closed and bounded...does this make sense?

5. yes, that makes sense. Thanks!

6. Originally Posted by sprinks13

2) suppose f:R->R is continuous on R and that lim(x->infinity)f = 0 and lim(x-> - infinity) f = 0. Prove that f is bounded and that it attains either a minimum or a maximum.

- attempt at a solution:
since I = R is an interval, and f is continuous on I, then f(I) is also an interval.
show that f(I) is bounded.
NO idea

thanks
I gave you the last one, so here are some hints

1. To prove that it there is a maximum or minimum apply Rolle's Theorem
2. To prove that it is bounded note that unboundedness is caused by either a simple discontinuity such as $\lim_{x\to a}f(x)\to\pm\infty$ or $f$ is unbounded as $|x|$ gets arbitrarily large. The latter is obviously not true and argue that the former is also false based on the continuity of $f$