# Find a continuous non 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 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?

• 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. – Severin Schraven Oct 13 '18 at 16:52
• 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 – Dominik Kutek Oct 13 '18 at 16:53
• We can just go with piece-wise linear function, that is $1$ on $[0, 1]$, and connects point $(1, 1)$ with $(2, A)$. – Jakobian Oct 13 '18 at 16:59

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 $$x=1$$.

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 $$\epsilon$$.