Triangle inequality for integrals, but for an arbitrary norm

238 Views Asked by At

Given an arbitrary norm on $\mathbb{R}^q$.

For a continous function $f: [a,b]\times\mathbb{R}^q \rightarrow \mathbb{R}^q$, I want to find out whether $$ \left\|\int_{a}^{b} f(x,y_1,...,y_q)\,\mathrm{d}x\right\| \le \int_{a}^{b} \| f(x,y_1,...,y_q)\|\,\mathrm{d}x$$ holds. I know that this is easily provable for the euclidean norm $\|\cdot\|_{2}$, but the proof I know involves the Cauchy-Schwarz inequality which is only available for induced norms. That is why I tried making use of the fact that all norms on $\mathbb{R}^q$ are equivalent, which means $$\exists\, \alpha,\beta>0\,\, \forall v\in\mathbb{R}^q:\alpha\|v\|\le\|v\|_{2}\le\beta\|v\|. $$ Using the triangle equality for $\|\cdot\|_{2}$, this yields $$ \alpha\left\|\int_{a}^{b} f\,\mathrm{d}x\right\| \le \left\|\int_{a}^{b} f\,\mathrm{d}x\right\|_{2}\le\int_{a}^{b} \| f\|_{2}\,\mathrm{d}x\le \beta\int_{a}^{b} \| f\|\,\mathrm{d}x,$$ but as there is no more information about $\alpha$ and $\beta$, I cannot conclude the above inequality.

I appreciate any help.