Find a sequence of functions that proves that two norms on $C^1 [0,1]$ are not equivalent. The problem is to find a sequence of $C^{1}[0,1]$-functions that converges with respect to the norm:
$$\left\lVert{f}\right\rVert_A=\max_{x\in[0,1]}\left\lvert{f(x)}\right\rvert+(\int_{0}^{1} \left\lvert{f'(t)}\right\rvert^2 dt)^\frac{1}{2},$$
but does not converge with respect to the standard norm in $C^{1}[0,1]$:
$$\left\lVert{f}\right\rVert_B=\max_{x\in[0,1]}\left\lvert{f(x)}\right\rvert+\max_{x\in[0,1]}\left\lvert{f'(x)}\right\rvert.$$
I have tried several simple functions, like $f_n=\log(x+\frac{1}{n})$, but none of them seemed to be suitable.
 A: I'll denote $\max_{x\in[0,1]}|f(x)|$ by $\|f\|_{\infty}$. Consider the sequence of functions $$f_n(x) = \frac{x^{n+1}}{n+1}.$$ We calculate the integrals of the squares of the derivatives:
$$\int_{0}^{1} |f_n'(t)|^2 \mathrm{d}t= \int_{0}^{1} t^{2n}\mathrm{d}t = \frac{1}{2n+1}.$$ Therefore, we obtain $$\|f_n\|_A = \frac{1}{n+1} + \sqrt{\frac{1}{2n+1}} $$ which shows that $f_n \to 0$ as $n \to \infty$ with respect to the $\|.\|_A$-norm.
For the $\|.\|_B$-norm: We have seen already that $\|f_n\|_{\infty} \to 0$, which shows that if the sequence $f_n$ converges in the $\|.\|_B$-norm, it has to converge against the constant $0$ function. But we have $$\|f_n\|_{B} = \frac{1}{n+1} + 1$$ which shows that $f_n$ does not converge to $0$ in the $\|.\|_{B}$-norm.
A: Take the sequence $(g_n)_{n\geq2}$ given by $g_n(x)=\sqrt{n}\cdot\max\{0,1-n^2|x-\frac{1}{2}|\}$, $0\leq x\leq1$. The plot of $g_n$ would be:
                                                    
Now consider $(f_n)_{n\geq2}$ defined by $f_n(x)=\int_0^xg_n(y)dy$, $0\leq x\leq1$, and observe that
$$
||f_n||_A=\frac{\sqrt{n}}{n^2}+\left(\int_0^1g_n(x)^2dx\right)^{1/2}\leq\frac{\sqrt{n}}{n^2}+\left(\int_{1/2-1/n^2}^{1/2+1/n^2}ndx\right)^{1/2}=\frac{\sqrt{n}}{n^2}+\left(\frac{2}{n}\right)^{1/2},
$$
which converge to $0$ as $n\to\infty$; but
$$
||f_n||_B\geq\max_{0\leq x\leq1}g_n(x)=\sqrt{n}\to\infty\text{ as }n\to\infty.
$$
