0
$\begingroup$

I am asked to find a continuous, not polynomial function so that $\max|f(x)|=1$ for $0\leq x \leq 1$ and $\max|f(x)|$ is as big as we want for $1\leq x \leq 2$. I've come up with

$$ f(x) = \left\{\begin{array}{lr} \frac{\cos(x)}{2}, & \text{for } x=0\\ \frac{1}{2-x}, & \text{for } 0< x\leq 2 \end{array}\right\} $$ Which is not very clever, what are some other examples of these kind of functions?

$\endgroup$
  • $\begingroup$ Your function is not defined at $x=2$. Also, such an example cannot exist as $[0,2]$ is a compact set and thus, every continuous function on it is bounded. $\endgroup$ – Severin Schraven Oct 13 '18 at 16:52
  • $\begingroup$ I really do believe, that by "max|f(x)| is as big as we want", he meant that if we want max|f(x)| to be some particular number A, we can achieve it $\endgroup$ – Dominik Kutek Oct 13 '18 at 16:53
  • $\begingroup$ We can just go with piece-wise linear function, that is $1$ on $[0, 1]$, and connects point $(1, 1)$ with $(2, A)$. $\endgroup$ – Jakobian Oct 13 '18 at 16:59
1
$\begingroup$

Your your proposed function doesn't work, since it's not defined at $x=2$.

Here's one that works . . .

Fix $M > 0$, and let $f$ be defined by $$ f(x)= \begin{cases} \sin(2\pi x)&\text{if $0\le x\le 1$}\\[4pt] M\sin(2\pi (x-1))&\text{if $1< x\le 2$}\\ \end{cases} $$ Notes:

The function $f$ above is not unbounded, but it's not possible for a continuous function on $[0,2]$ to be unbounded.

However on $[0,1]$, the maximum absolute value of $f$ is $1$, and on $[1,2]$, the maximum absolute value of $f$ is $M$, which can be made arbitrarily large.

The continuity of $f$ is verified since both pieces "join" at $x=1$.

$\endgroup$
0
$\begingroup$

You don't need the piecewise definition. You could simply do $$f(x)=\frac{1}{(2+\epsilon-x)^p}$$ for some small $\epsilon>0$ and $p\geq1$. Then $$\max_{0\leq x\leq1}f(x)=\frac{1}{(1+\epsilon)^p}<1$$ and $$\max_{1\leq x\leq2}f(x)=\frac{1}{\epsilon^p}$$ which can be made arbitrarily large by decreasing $\epsilon$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.