Exercise :
Over $C[a,b]$, we define the norm : $$\| f \|_2 = \int_a^b f(x)^2\mathrm{d}x$$ Show that $\| \cdot \|_2$ indeed defines a norm.
Attempt :
It's easy to prove $\| \cdot \|_2 \geq 0$, as :
$$\| f \|_2 = \int_a^b f(x)^2\mathrm{d}x \geq 0 \; \forall f \in C[a,b]$$
with $\| f \|_2 = 0$ for :
$$\|f\|_2 = 0 \Leftrightarrow \int_a^b f(x)^2\mathrm{d}x = 0 \Leftrightarrow f(x) = 0 $$
Now, it is :
$$\| \lambda f\|_2 = \int_a^b(\lambda f(x))^2\mathrm{d}x =\int_a^b \lambda^2f(x)^2\mathrm{d}x = \lambda^2\int_a^bf(x)^2\mathrm{d}x = \lambda^2\|f\|_2 \neq \lambda \|f\|$$
What's the issue here ? It seems that $\|\lambda f\|_2 \neq \lambda \|f\|_2$.
Also, for the triangle inequality, it is :
$$\|f+g\|_2 = \int_a^b(f+g)^2(x)\mathrm{d}x = \int_a^b(f(x)^2 + 2f(x)g(x) + g(x)^2)\mathrm{d}x$$
$$= \int_a^b f(x)^2\mathrm{d}x + \int_a^b 2f(x)g(x)\mathrm{d}x + \int_a^b g(x)^2\mathrm{d}x = \|f\|_2 + \int_a^b2f(x)g(x)\mathrm{d}x+\|g\|_2 $$
Now, I fail to see why would this be $\leq \|f\|_2 + \|g\|_2$.
Please assist me in understanding and solving the issues above.