Using Laplace transforms I created a proof that cos(t) is not bounded above. I cannot find where the error is even though the proof doesn't use very complicated theorems. I am not so familiar with the limitations of Laplace transforms so I suspect that the error is at the start or the end where I use Laplace.

If cos(t) is not bounded above then for any c>1 there exists some t such that $\displaystyle cos(t)=c$

Taking the Laplace transform of the equation

$\displaystyle \mathcal{L} \left\{ cos(t) \right\} = \mathcal{L} \left\{ c \right\}$

$\displaystyle \frac{s}{s^2+1} = \frac{c}{s}$

Assuming s is not equal to zero

$\displaystyle \frac{s^2}{s^2+1} = c$

There is no solution for real s, instead consider a purely complex value $\displaystyle s=xi$, where $\displaystyle x \in R \backslash 0$

$\displaystyle \frac{-x^2}{-x^2+1} = c$

Plotting $\displaystyle \frac{x^2}{x^2-1}$ gives the following graph:

As can be seen from the graph, the function is continuous for x>1

Now using the following limits:

$\displaystyle \lim_{x \to 1^+}\frac{x^2}{x^2-1} =\infty$

and

$\displaystyle \lim_{x \to \infty}\frac{x^2}{x^2-1} =1$

Since $\displaystyle c \in (1,\infty)$ then by the intermediate value theorem there exists some $\displaystyle x_0 \in (1,\infty)$ such that $\displaystyle \frac{x^2}{x^2-1} =c$

Therefore a solution in the Laplace domain $\displaystyle s_0=x_0i$ exists and this implies that a solution in the time domain exists for some $\displaystyle t>0$

Is anyone able to spot the error in my reasoning? I didn't properly prove that the function was continuous when applying the intermediate value theorem but I don't think this is the blunder.