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

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 $$\sqrt{\int_0^1|f(t)+g(t)|^2dt}\le\sqrt{\int_0^1|f(t)|^2dt}+\sqrt{\int_0^1|g(t)|^2dt}$$

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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?

Answer 1

Use Minkowski's inequality: $$\int_0^1|f+g|^2\le\int_0^1|f|^2+\int_0^1|g|^2+2\int_0^1|fg|\le \int_0^1|f|^2+\int_0^1|g|^2+2\sqrt{\int_0^1|f|^2 \int_0^1 |g|^2}.$$

An edit to reflect another question from the author

If you consider complex-valued functions, then you define your inner product as $$(f,g)=\int_0^1 f(t) \overline{g(t)}dt.$$ Note, that inner product is not bilinear but linear in one variable and semilinear in another. It's easy to show that this expression satisfies the axioms for inner product: $$(\alpha f,\beta g) = \alpha \bar \beta (f,g), $$ $$ ( f, g+h) =( f, g )+( f, h), $$ $$ ( f+h, g) =( f, g )+( h, g), $$ $$(f,f)\ge 0,$$ $$(f,f)=0\iff f=0.$$ and that $\|f\|_{2} = \sqrt{(f,f)}$.

If we want prove Minkowski's inequality for our case, we write $$\int_0^1|f+g|^2=(f+g,f+g)=\int_0^1(f+g)(\bar f +\bar g)=\int_0^1 f \bar f + \int_0^1 (g \bar f +f\bar g)+\int_0^1 g \bar g $$

$$=\|f\|_2^2+\|g\|_2^2+ (f,g)+(g,f)$$ As $|(f,g)|=|(g,f)|\le \|f\|_2\|g\|_2$ by the Cauchy-Schwarz inequality, because it works for complex-valued functions (see its proof; note that the proof doesn't use the triangle inequality), so we can continue

$$\le \|f\|_2^2+\|g\|_2^2+ 2 \|f\|_2\|g\|_2, $$ or in other words $$\|f+g\|_2^2\le (\|f\|_2+\|g\|_2)^2$$ or $$\|f+g\|_2 \le \|f\|_2+\|g\|_2 $$

Answer 2

If by $C_\mathbb{R}([0,1],\mathbb{C})$ you mean the complex valued continuous functions on $[0,1]$ viewed as a real vector space then you only need to note that this space is the same as $C([0,1],\mathbb{R}^2)$ as functions spaces and as vector spaces over $\mathbb{R}$. Then the inner product you want is

$$ (f,g)= \int_0^1 f(x)\cdot g(x) dx = \int_0^1 f_1(x)g_1(x)+f_2(x)g_2(x) dx, $$ where $f=(f_1,f_2)$ and $g=(g_1,g_2)$. C-S in the space follows directly from C-S in $\mathbb{R}^2$ and $L^2$ since $|f\cdot g|\leq |f| |g|$ so that

$$ |(f,g)| \leq \int_0^1 |f(x)||g(x)| dx \leq \left( \int_0^1 |f(x)|^2dx \right)^{1/2}\left( \int_0^1 |g(x)|^2dx \right)^{1/2}. $$

The triangle inequality also follows from this since

$$ \| f+g\|^2 =(f+g,f+g) = \| f\|^2 +2(f,g) +\| g\|^2 \leq \| f\|^2 +2\| f\|\| g\| +\| g\|^2 =(\| f\|+ \| g\|)^2. $$

Answer 3

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.

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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: $$ \|f\|_2^2=\langle f,f\rangle\tag{6} $$