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I`m currently having some issues with a seemingly innocent problem. I would like to show that

$$\Bigg|\Bigg|\int_\mathbb{R}\begin{pmatrix}A(x)\\B(x)\end{pmatrix}dx\Bigg|\Bigg|_2 \leq \int_{\mathbb{R}}\Bigg|\Bigg|\begin{pmatrix}A(x)\\B(x)\end{pmatrix}\Bigg|\Bigg|_2dx$$

Where $A(x),B(x) \in L^2(\mathbb{R})$ and the two norm is defined as $$\Bigg|\Bigg|\begin{pmatrix}A(x)\\B(x)\end{pmatrix}\Bigg|\Bigg|_2=\sqrt{|A(x)|^2+|B(x)|^2}$$

I've asked around and people have tended to say "that's very simple" and then spent half an hour staring at it. I've tried plugging stuff in and it seems to hold but I do need a proof. Any help would be much appreciated!

Thanks in advance

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Since $\lVert\cdot\rVert_2$ is a norm, isn't it essentially the triangle inequality? (google "integral triangle inequality", for instance) – Clement C. Feb 15 at 14:59
Your $2$-norm takes two things as argument, as you've defined it. So what does it mean to take the $2$-norm of an integral which is just one argument? – Gregory Grant Feb 15 at 15:02
@GregoryGrant Isn't the integral of a vector a vector? (i.e., it's the component-wise integral) – Clement C. Feb 15 at 15:03
@GregoryGrant the integral of a vector is a vector. One can see this if they write out the vector as a sum. – Oliver Brace Feb 15 at 15:06
@ClementC. Oh you mean like ${{\int_{\Bbb R}A(x)dx}\choose{\int_{\Bbb R}B(x)dx}}$? – Gregory Grant Feb 15 at 15:07
up vote 6 down vote accepted

It's very simple. hehe... (Note you actually want to assume $A,B\in L^1(\Bbb R)$, not $L^2$.)

Edit: Morally the same argument works for a Banach-space valued function; see Below.

To make things easier to type I'm going to revise the notation. Suppose that $f:\Bbb R\to\Bbb R^2$; we want to show that $$\left|\left|\int f(x)\,dx\right|\right|_2\le\int||f(x)||_2\,dx.$$Let $$v=\int f(x)\,dx\in\Bbb R^2.$$Then $$||v||_2^2=v\cdot v=v\cdot\int f(x)\,dx=\int v\cdot f(x)\,dx\le\int||v||_2||f(x)||_2\,dx=||v||_2\int||f(x)||_2\,dx.$$Divide by $||v||_2$: $$||v||_2\le\int||f(x)||_2\,dx.$$

Below One can give a similar argument if $f:S\to X$ where $X$ is a Banach space and $\mu$ is a measure on $S$. Let $$v=\int_Sf(t)\,d\mu(t)\in X.$$Suppose $\Lambda\in X^*$ and $||\Lambda||=1$. Then $$\Lambda v=\int\Lambda f(t)\,d\mu(t)\le\int||\Lambda||_{X^*}||f(t)||_X\,d\mu(t)=\int||f(t)||\,d\mu(t).$$Since this holds for every such $\Lambda$, Hahn-Banach shows that $||v||\le\int||f||\,d\mu$.

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Thank you. You've made my project a bit more bearable. – Oliver Brace Feb 15 at 16:08

To give a more general picture, that does not use the special form of the $2$-norm being induced by the scalar product, we will show:

Proposition. Suppose $X$ is a Banach space, $(S,\mathcal A, \mu)$ a measure space, and $f \colon S \to X$ is integrable. Then we have $$ \def\norm#1{\left\|#1\right\|}\norm{\int_S f \, d\mu} \le \int_S \norm {f(s)} \, d\mu(s) $$

Proof. We use the definition of the integral. If $f = \sum_i x_i\chi_{A_i}$ is a simple function, where the $A_i$ are disjoint, then $\norm{f(s)} = \sum_i \norm{x_i} \chi_{A_i}(s)$ and hence \begin{align*} \norm{\int_S x_i \chi_{A_i}d\mu} &= \norm{\sum_i \mu(A_i)x_i}\\ &\le \sum_i \mu(A_i)\norm{x_i}\\ &= \int_S \sum_{i}\norm{x_i}\chi_{A_i} d\mu\\ &= \int_S \norm{f(s)}\, d\mu(s) \end{align*} If $f$ is integrable, choose simple functions $f_n$ such that $f_n \to f$ almost everywhere and $\lim_n\int_S f \, d\mu = \int_S f_n \, d\mu$ we have \begin{align*} \norm{\int_S f\, d\mu} &= \lim_n \norm{\int_S f_n\, d\mu}\\ &\le \lim_n \int_S \norm{f_n(s)}\, d\mu(s)\\ &= \int_S \norm{f(s)}\, d\mu(s) \end{align*}

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Or one can give an argument in the Banach-space case that's precisely analogous to the argument using the fact that the $2$-norm is induced by an inner product... see edit to my answer. – David C. Ullrich Feb 15 at 15:37
Thanks, great addition. – martini Feb 15 at 15:40
Both arguments are worth knowing; yours is more elementary, while I think it's fair to say that mine is simpler. (And there are related situations where someone who only knows one would be better off knowing the other...) – David C. Ullrich Feb 15 at 15:45
Thank you but the other answer is more specific to my problem. – Oliver Brace Feb 15 at 16:09

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