Suppose $f :[a,b]\to \mathbb{R}^n$ is given by $f(t)=\langle f_1(t), \ldots, f_n(t)\rangle$, where each of the $n$ component functions is integrable over $[a,b]$. I think the following inequality

$$\left|\int _a^b f(t) \, dt\right|\leq\int_a^b\left|f(t)\right| \, dt$$

still hold but I don't know how to prove it. Any idea?


  • $\begingroup$ I changed $<f\cdots f>$ to $\langle f\cdots f\rangle$. That is standard usage. $\endgroup$ Nov 19 '12 at 22:46
  • $\begingroup$ May I ask how you do it? $\endgroup$
    – KWO
    Nov 19 '12 at 22:48
  • $\begingroup$ I coded the second expression above as \langle f\cdots f\rangle. If you right-click on the expression as it is rendered, you will see a menu that says "Show Math As". Then choose "TeX Commands" and you'll see it. $\endgroup$ Nov 19 '12 at 22:53
  • $\begingroup$ . . . . also, if you click on "edit" on your question, you'll see what's there. $\endgroup$ Nov 19 '12 at 22:54

The inequality you want is a generalization of the triangle inequality. You can either use the triangle inequality at the level of the Riemman sums: $$ \left|\int_a^bf(t)dt\right|=\left|\lim_n \sum_{k=1}^n f(t_{n,k})(t_{n,k}-t_{n,k-1})\right|=\lim_n \left|\sum_{k=1}^n f(t_{n,k})(t_{n,k}-t_{n,k-1})\right|\\ \leq\limsup_n \sum_{k=1}^n |f(t_{n,k})|(t_{n,k}-t_{n,k-1})=\int_a^b|f(t)|dt $$ (where the last equality is justified by the fact that $|f(t)|$ is integrable, being a composition of a continuous function and integrable functions).

Or you can calculate $$ \left|\int_a^bf(t)dt\right|^2=\left\langle\int_a^bf(t)dt,\int_a^bf(t)dt\right\rangle=\int_a^b\int_a^b\langle f(t),f(s)\rangle\,dtds\\ \leq\int_a^b\int_a^b| f(t)|\,|f(s)|\,dtds=\left(\int_a^b|f(t)|dt\right)^2 $$ (where you would still need to justify that the integral of the inner product is the inner product of the integrals, again by the linearity and continuity of the inner product).

  • $\begingroup$ Thanks, I am still thinking about your second proof, it is much elegant but I am still trying to work out the details. $\endgroup$
    – KWO
    Nov 19 '12 at 23:06
  • 1
    $\begingroup$ To prove that you can take the integral out of the inner product, just work with one integral on one side and a fixed vector on the other; that will simplify computations and you still get the result you need. $\endgroup$ Nov 19 '12 at 23:08

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