TITLE: Does $\|\cdot\|_2:C_\mathbb R([0,1],\mathbb C)\to\mathbb R:f\mapsto\sqrt{\int_0^1|f(t)|^2dt}$ come from any inner product?
QUESTION [8 upvotes]: I'm  trying to show $\|\cdot\|_2$ is a norm on the $\mathbb C$-vector space $C([0,1],\mathbb C)$ where $$\|\cdot\|_2:C([0,1],\mathbb C)\to\mathbb R:f\mapsto\sqrt{\int_0^1|f(t)|^2dt}$$
I've stuck in course of showing triangle inequality. 
I've to show that $$\sqrt{\int_0^1|f(t)+g(t)|^2dt}\le\sqrt{\int_0^1|f(t)|^2dt}+\sqrt{\int_0^1|g(t)|^2dt}$$

Edit: I've already received an answer using Minkowski's Inequality. Can't we conclude it by Cauchy-Schwarz even when the base field is $\mathbb R$ instead of $\mathbb C$? I can see for the vector space $C([0,1],\mathbb R)$ over $\mathbb R$ the triangle inequality follows from Cauchy-Schwarz considering the inner product $$(,):C_\mathbb R([0,1],\mathbb R)\times C_\mathbb R([0,1],\mathbb R)\to\mathbb R:(f,g)\mapsto\int_0^1f(t)g(t)dt.$$ But if we consider the linear space $C_\mathbb R([0,1],\mathbb C)$ then not necessarily $\int_0^1f(t)g(t)dt\in\mathbb R$ neither $\int_0^1f(t)\overline{g(t)}dt\in\mathbb R$. However $$(,):C_\mathbb R([0,1],\mathbb C)\times C_\mathbb R([0,1],\mathbb C)\to\mathbb R:(f,g)\mapsto\int_0^1|f(t)g(t)|dt$$ is not an inner product for it doesn't obey $(f,g+h)=(f,g)+(f,h).$ 

How is it possible to solve it by Cauchy-Schwarz? 
Does $\|\cdot\|_2:C_\mathbb R([0,1],\mathbb C)\to\mathbb R:f\mapsto\sqrt{\int_0^1|f(t)|^2dt}$ even come from any inner product?

REPLY [2 votes]: The triangle inequality for the $L^2$ norm is precisely Minkowski's inequality for the $L^2$ norm.
One simple proof of Minkowski's inequality uses duality. That is,
$$
\|f\|_2=\sup_{\|h\|_2=1}\,\left|\,\int_0^1f(x)h(x)\,\mathrm{d}x\,\right|\tag{1}
$$
Using $(1)$, we get
$$
\begin{align}
\|f+g\|_2
&=\sup_{\substack{\|h_1\|_2=1\\\|h_2\|_2=1\\h_1=h_2}}\,\left|\,\int_0^1\big(f(x)h_1(x)+g(x)h_2(x)\big)\,\mathrm{d}x\,\right|\\
&\le\sup_{\substack{\|h_1\|_2=1\\\|h_2\|_2=1}}\,\left|\,\int_0^1\big(f(x)h_1(x)+g(x)h_2(x)\big)\,\mathrm{d}x\,\right|\\[12pt]
&=\|f\|_2+\|g\|_2\tag{2}
\end{align}
$$
The inequality follows because the second $\sup$ is being taken over a larger set of $h$s.
To prove $(1)$, we use Cauchy-Schwarz to show
$$
\sup_{\|h\|_2=1}\left|\,\int_0^1f(x)h(x)\,\mathrm{d}x\,\right|\le\|f\|_2\tag{3}
$$
Since $\|f\|_2^2=\int_0^1f(x)\overline{f(x)}\,\mathrm{d}x$, if $h(x)=\overline{f(x)}/\|f\|_2$, then $\|h\|_2=1$ and
$$
\int_0^1f(x)h(x)\,\mathrm{d}x=\|f\|_2\tag{4}
$$
Thus, $(1)$ is true.

As alluded to above, the complex inner-product
$$
\langle f,g\rangle=\int_0^1f(x)\overline{g(x)}\,\mathrm{d}x\tag{5}
$$
leads to the $L^2$ norm:
$$
\|f\|_2^2=\langle f,f\rangle\tag{6}
$$