A common example of an illposed problem is the following differentiation example:
Let $y(t)\in C^1([0,1])$ with $y(0)=0$. We consider the problem of determining $x(t) = y'(t)$ and show why it is not wellposed due to stability issues. For example, let $y(t) = t$. Then $x(t)=1$. Then let us add a small perturbation $\tilde y(t) = t + \epsilon \sin(kt)$ for $\epsilon >0$ arbitrarily small and $k>0$ arbitrarily large. The perturbation in the input data $y(t)$ is small. The perturbation in the output data $x(t)$ is not because, if for example we use the supremum norm, $$\max_{t\in [0,1]} | \tilde x - x | =\max_{t\in [0,1]} | 1+k\epsilon \cos (kt)-1 | = k\epsilon$$ which is arbitrarily large. Now here is where I got confused. Why is the problem wellposed when using the $C^1$ norm, instead of the $C^0$ norm? The error is then $k^2\epsilon$ which is still arbitrarily large?!
The problem is well-posed if the pre-image space is $C^1$. Then the $C^0$-error is still $k\epsilon$, but this is equal to $\|y-\tilde y\|_{C^1}$, hence bounded against the error in the data.