I am reading a book about functional analysis and I found the following example proving that some sequence converges weakly* but not weakly.
Let $X=C[-1,1]$ be the space of continuous functions and $$\rho_n(t)=\left\{ \begin{array}{ccl} n-n^2|t| &\text{if}& -\frac{1}{n} \leq t \leq \frac{1}{n} \\ 0 &\text{otherwise}& \end{array} \right.$$ We consider the seequence functionals $f_n:X \to \mathbb{C}$ defined by $$f_n(\phi)=\int_{-1}^1\phi(t)\rho_n(t)\text{d}t, \qquad \phi\in C[-1,1].$$ We claim that $f_n$ weak^* converges to $f_0$ defined by $f_0(\phi)=\phi(0)$. Indeed, \begin{equation*} \begin{split} |f_n(\phi)-f_0(\phi)|& =\left|\int_{-1}^1\phi(t)\rho_n(t)\text{d}t-\int_{-1}^1\phi(0)\rho_n(t)\text{d}t\right| \\ & \leq \int_{-1}^1 |\phi(t)-\phi(0)|\rho_n(t)\text{d}t=\int_{-1/n}^{1/n} |\phi(t)-\phi(0)|\rho_n(t)\text{d}t \\ & \leq \max\{|\phi(t)-\phi(0)| : -1/n\leq t \leq 1/n\}. \end{split} \end{equation*}
My question is: What happened in the last inequality? I know that given a Riemann-integrable function we have that $$\int_a^b f(x)\text{d}x\leq (b-a)\max\{|f(t)| : t\in[a,b]\}$$ but I do not know how to apply similar results here.
The expression $|\phi(t)-\phi(0)|$ is "pulled out" the integral sign and is majorized by the obvious $\max\{|\phi(t)-\phi(0)|: |t|\leq 1/n\}$, and you are left with the integral $$\int_{-1/n}^{1/n}\rho_n(t)\,dt = 1$$
so that is what happened in the last inequality. (The first (or second) question you should have asked yourself is this: whatever happened to $\rho_n$?)