Show that the functional is continuous

Question

Task:

Check if the functional F: $$F(f) = \int_{-1}^1 f(t) sgn(t) dt.$$is continuous on the space $\mathbb E=L_2(-1, 1) $:

Solution:

I need to find $M>0$ and show the inequality: $$\vert F(f) \vert \le M (\int_{-1}^1 \vert f(t)\vert^2 dt)^{1/2}.$$

I started from $$ \vert F(f) \vert = \vert \int_{-1}^1 f(t) sgn(t) dt \vert$$

But actually I do not know how to evaluate this function.

I know that somewhere I need to use the Hölder's inequality: $$ \int_{-1}^1 \vert f(t) g(t) \vert dt \lt (\vert f(t) \vert^2 dt)^{1/2} (\vert g(t) \vert^2 dt)^{1/2},$$

Answer 1

You can apply Hölder with $g=1$, i.e. $$ \left| \int_{-1}^1 f(t) \, \text{sgn}(t) \, \mathrm dt \right| \leq \int_{-1}^1 |f(t)|\cdot 1 \, \mathrm dt \leq \left(\int_{-1}^1 |f(t)|^2 \, \mathrm dt\right)^{1/2}\left(\int_{-1}^1 1^2 \, \mathrm dt\right)^{1/2} = \sqrt 2\left(\int_{-1}^1 |f(t)|^2 \, \mathrm dt\right)^{1/2}. $$