counterexample for ill posedness of the laplace equation Consider the wave equation with initial data:
$$u_{tt}(t,x) + u_{xx}(t,x) = 0$$
$$u(0,x) = u_0(x)$$
$$u_t(0,x) = u_1(x)$$ 
Hadamard showed that this problem is ill-posed: there exist large solutions with arbitrarily small initial data. For instance, if we take $u(t,x) = a_{\omega} \sinh(\omega t) \sin(\omega x)$, then $u_0(x) = 0$ and $u_1(x) = a_{\omega} \omega \sin (\omega x)$, then we can make $u(t,x)$ grow arbitrarily fast while keeping $u_0$ and $u_1$ small. 
Tweaking this construction, it is not hard to see that for any given $k$ and $\epsilon > 0$, we can construct initial data such that 
$$
||u_0||_{\infty} + ||u_0^{(1)}||_{\infty} + \ldots + ||u_0^{(k)}||_{\infty} + ||u_1||_{\infty} + ||u_1^{(1)}||_{\infty} + \ldots + ||u_1^{(k)}||_{\infty} < \epsilon
$$
and $||u(\epsilon,x)||_{\infty}  > \frac{1}{\epsilon}$. This can be interpreted as saying that the problem is ill-posed even in a Holder sense.  
My question is: can one construct an example of a solution $u(t,x)$ with initial data $u_0(x)$ and $u_1(x)$ such that 
$$\sum_{i=0}^{\infty} ||u_0^{(i)}||_{\infty} + ||u_1^{(i)}||_{\infty} < \epsilon$$
while $||u(\epsilon,x)||_{\infty}  > \frac{1}{\epsilon}$? This is exercise 26 in http://www.math.mcgill.ca/gantumur/math580f11/downloads/notes2.pdf, which suggests that it should be possible. An explicit construction may be difficult, in which case I would be happy with an abstract argument for existence. 
 A: I am somewhat doubtful of the claim. Here's why. Let us suppose the weaker inequality that for any $k$
$$ |\partial_x^k u_0| < \epsilon, |\partial_x^k u_1| < \epsilon. $$
Now, let us consider a solution $u$ to the Laplace equation with indicated boundary values. We have that along $t = 0$
$$ |\partial_x^k u| < \epsilon, |\partial_x^k \partial_t u| < \epsilon. $$
Using the equation we have
$$ |\partial_x^k \partial_t^2 u| = |\partial_x^{k+2} u| < \epsilon. $$
And by induction
$$ |\partial_x^k \partial_t^{2l} u| = |\partial_x^{k+2l} u| < \epsilon. $$
and
$$ |\partial_x^k \partial_t^{2l+1} u| = |\partial_x^{k+2l} \partial_t u| < \epsilon. $$
This implies that every single derivative, spatial or temporal is bounded by $\epsilon$. The practical implication of this is: let 
$$ a_{kl} = \partial_x^k \partial_t^l u(0,0) $$
The formal power series
$$ U(t,x) = \sum_{k,l = 0}^{\infty} \frac{1}{k!l!} a_{kl} x^k t^l $$
has infinite radius of convergence; the function that it defines is a solution, so we can identify $U$ with $u$. 
Now, if you look at $t = \epsilon, x = 0$ of the power series, you see that you have the sum 
$$ U(\epsilon,0) = \sum_{l = 0}^{\infty} \frac{1}{l!} a_{0l} \epsilon^l$$ 
which we can bound pretty trivially by 
$$ (\sup_l |a_{0l}|) \exp(\epsilon) < 2 \epsilon $$
The above argument is clearly independent of where we do the Taylor series: instead of taking the series relative to the origin, we do it relative to $(0,x_0)$ and this tells us that the desired inequality is very far from being true. 

The above is basically a quantitative version of the theorem of Cauchy-Kowalevski, and which can be easily extended to initial data in Gevrey classes; we are using here that if a smooth function is such that all its derivatives are bounded by a fixed constant, the function must be real analytic. 
