Let $X_1$ respectively $X_2$ denote the space $C [a, b]$, $a <b$ with norm:
$$\Vert x\Vert_1=\int_a^b \vert x(t)\vert dt; $$$$\Vert x\Vert_2=\left(\int_a^b \vert x(t)^2\vert dt\right)^{\frac{1}{2}}.$$
Using Bunyakovsky-Schwarz inequality for $x\in C[a,b]$ we have:
$$\Vert x\Vert _1 = \int_a^b\vert 1\cdot x(t)\vert\leq\left(\int_a^b dt\right)^{\frac{1}{2}}\left(\int_a^b\vert x(t)\vert^2 dt \right)^{\frac{1}{2}}$$
$\Rightarrow \Vert x\Vert _1 \leq M \Vert x \Vert _2, $ where $M=\sqrt{b-a}.$
On the other hand, the function $x_n$ defined below
$$x_n(t)=\left\{\begin{array}{cc}0, &\mbox{ if } a \leq t\leq b-\frac{1}{n} \\n, & \mbox{ if } b-\frac{1}{2n}\leq t\leq b \\\mathrm{affin}, & \mbox{ if } b-\frac{1}{n}\leq t\leq b- \frac{1}{2n}\end{array}\right.$$
With $t$ danote a $x$ axis and with $x$ danote $y$ axis.
Now we have: $$\Vert x_n\Vert_2^2\geq\frac{n}{2}\Rightarrow \Vert x_n\Vert _2\geq\sqrt\frac{n}{2}.$$ I understand this.
But I dont understand this: As you know that:
$$\Vert x_n\Vert _1=\frac{3}{4}$$
Thanks for your help and your attention.
