Let $X =C_{L^2}([-1,1], \mathbb{R})$ which is the space of continuous functions on the intervall [-1,1] which map to the real numbers, equipped with the $L^2$ inner product, and a functional $F:X \to \mathbb{R}$ defined by $$F(x) = \int_{0}^{1}x(t)dt.$$ Show that $F \in C_{L^2}([-1,1], \mathbb{R})^*$ and calculate $||F||_{X^*}$.
I know that for $F$ to be an element of the dual space of $X$, $F$ must be linear and continuous. $F$ is linear because of the linearity of the integral and for it to be continuous it must be bounded. I tried to show that $F$ is bounded but I ran into problems because the integral limits are not the same as the limits of $X$. My idea was: $$|F(x)|=\Bigg|\int_{0}^{1}x(t)dt + \int_{-1}^{0}x(t)dt - \int_{-1}^{0}x(t)dt\Bigg|=\Bigg|\int_{-1}^{1}x(t)dt-\int_{-1}^{0}x(t)dt\Bigg|.$$ Then for a $y \in X$ defined by $y \equiv 1$ it follows: $$\Bigg|\int_{-1}^{1}x(t)dt-\int_{-1}^{0}x(t)dt\Bigg|=\Bigg|\langle x,y\rangle_{L^2[-1,1]} - \int_{-1}^{0}x(t)dt\Bigg|.$$ I know that by using Cauchy-Schwarz inequality that $$\langle x,y\rangle_{L^2[-1,1]} \le \Bigg(\int_{-1}^{1}x^2dt\Bigg)^{\frac{1}{2}}\Bigg(\int_{-1}^{1}y^2dt\Bigg)^{\frac{1}{2}}=\sqrt2\cdot||x||_{L^2[-1,1]}$$ Now if I could get rid of the integral in the absolute value I would be done but I don´t know how to progress from this point. I also have problems with calculating $||F||_{X^*}$ since $X$ isn´t a Hilbert space and therefore I can´t use the Riesz representation theorem. Please correct my where I´m wrong, help would be very much appreciated! Also I hope that everything is understandable since this is my first post and english isn´t my first language. Thank you very much in advance!