Continuity of a functional with respect to two different norms Let $J$ be a functional defined on $E = C^1[a,b]$ by
$$J(y) = \int_a^b \sqrt{1 + (y^{\prime}(x))^2} \, dx.$$
Define the following two norms on $E$:
$$\|y\|_{\infty} = \max_{a\leq x\leq b} |y(x)|$$
$$\|y\|_{\ast} = \|y\|_{\infty} + \|y^{\prime}\|_{\infty}.$$
Then I'm to show that $J$ is continuous on $(E, ||\cdot||_{\ast})$ but not $(E, \|\cdot \|_{\infty})$.
First question. As far as I can tell, the first proof should be routine. Start with $(y_n)$ cauchy and choose $\epsilon$, then pick $N$ such that
$$||y_n - y_m||_{\ast} < \epsilon$$
for all $n,m > N$. Then find $\delta$ such that
$$|J(y_n) - J(y_m)| < \delta.$$
However, I'm completely blanking on how to manipulate the expanded expression
$$\left|\int_a^b \sqrt{1 - y_n^{\prime}}dx - \int_a^b \sqrt{1 - y_m^{\prime}}dx\right|.$$
Second question: I know that all I need to do to prove that $J$ is discontinuous in the other norm is provide a counterexample, but how on earth do I generate one? Am I just supposed to intuit the form of a counterexample or is there some strategy for producing one here?
 A: To show continuity in the norm $\|\cdot\|_*$ you can take advantage of the uniform continuity of the function $f(z) = \sqrt{1+z^2}$ on a compact set.
Let $\epsilon > 0$. Since $y_1(x) \in C^1$ then $y_1'[a,b]$ is a compact set as is $$B = \{ z \in \mathbb{R} : |z-w| \le 1 \text{ for some } w \in y_1'[a,b] \}.$$ $f$ is uniformly continuous on $B$ which means there is a $1 > \delta > 0$ such that if $z,w \in B$ and $|z-w| < \delta$ then $|f(z)-f(w)| < \epsilon$.
Now let $y_2(x)$ be such that $$\|y_1 - y_2 \|_* < \delta.$$ Since $\delta < 1$ we have $y_2'(x) \in B$ for all $x \in [a,b]$, since $|y_1'(x) - y_2'(x)| < \delta < 1$ for all $x$. Hence $$\left|\int_a^b \sqrt{1+y_1'(x)^2}dx - \int_a^b \sqrt{1+y_2'(x)^2}dx\right| < (b-a) \epsilon$$

As for the counter example, the norm $\|\cdot\|_\infty$ gives no control over the derivative, whereas the norm $\|\cdot\|_*$ does. So now consider the function $y_1(x)=0$ for which $J(y_1) = b-a$. Now we wish to construct a sequence of functions $f_n$ that converge to $y_1$.
Notice that $J$ is the arc length of the curve. Try to find a counter example with $\sin(x)$ such as $$f_n(x)=\frac{1}{\sqrt{n}}\sin(nx)$$
Here the amplitude of $f_n$ goes to zero, so it converges in the infinity norm. However, if the frequencies of $\sin(nx)$ increase fast enough (maybe with $n^2x$ in the argument even) then the arc length will get very large.
