0
$\begingroup$

I have the norm on the set of continuous functions from $s$ to $t$. Let $g$ and $h$ be in this set and define the dot product as $$\int_s^t g(x)h(x) \mathrm{dx}$$ and norm by $$ \|g\|_2 = \sqrt{g \cdot g} = \left (\int_s^t \lvert g(x)\rvert^2 \mathrm{dx} \right)^{1/2}$$ Show that it is a norm.

For the first part I have $$\|g\|_2 = \sqrt{g \cdot g} = \left (\int_s^t \lvert g(x)\rvert^2 \mathrm{dx} \right)^{1/2}$$ and clearly $\lvert g(x)\rvert$ is greater than $0$, hence $\|g\|_2 \ge 0$. Suppose $\|g\|_2 = 0$ I am having a little bit of trouble showing that $g$ must be $0$.

For the second part I have $\|c\cdot g(x)\|_2 = \sqrt{(c \cdot g(x))\cdot (c \cdot g(x))} = \sqrt{c^2 \cdot g \cdot g} = \sqrt{c^2}\sqrt{g \cdot g} = c \left (\int_s^t \lvert g(x)\rvert^2 \mathrm{dx} \right)^{1/2} = \|c\|\|g(x)\|$

I need help showing the triangle inequality for this norm to be true. I believe I am allowed to prove the triangle inequality for its' square and that implies the original triangle inequality is true.

I have $$\|g + h\|_2^2 = \|g\|_2^2 + \|h\|_2^2 + 2\cdot (g \cdot h) \\=\|g\|_2^2 + \|h\|_2^2 + 2\cdot\int_s^t g(x) h(x)\mathrm{dx}$$

I am not sure where to go from here. Any help would be appreciated

$\endgroup$
0

2 Answers 2

Reset to default
1
$\begingroup$

Let's show that $\lVert \cdot \rVert_2$ is a norm. We will be using the properties of the dot product. I will note $E = \mathcal{C}^0([s,t],\mathbb{R})$ the real vector space of continuous functions from $[s,t]$ to $\mathbb{R}$, and $(\cdot | \cdot)$ the dot product for more clarity. As you saw, $\lVert \cdot \rVert_2$ is well defined.

Homogeneity : Let $c \in \mathbb{R}$ and $g \in E$ then

$$\lVert cg \rVert_2^2=(c g|cg)=c^2\lVert g \rVert_2^2$$ Hence, $$\lVert cg \rVert_2 = |c|\lVert g \rVert_2$$

Positive definiteness : for all $g \in E$, if $\lVert g \rVert_2 =0$ then it means $\lVert g \rVert_2^2 =0$ so $(g|g)=0$. But $(\cdot|\cdot)$ is a dot product so it verifies positive definiteness, thus $(g|g)=0$ implies $g=0$.

Triangular inequality : Let $f,g \in E$. By the Cauchy-Schwarz inequality we have:

$$(f|g)^2 \leq (f|f)\,(g|g) $$

Hence,

$$(f|g) \leq |(f|g)|\leq\lVert f\rVert_2 \lVert g\rVert_2 $$

Moreover,

$$\lVert f+g\rVert_2^2 = \lVert f\rVert_2^2+\lVert g\rVert_2^2+2(f|g) \leq \lVert f\rVert_2^2+\lVert g\rVert_2^2+2\lVert f\rVert_2 \lVert g \rVert_2$$

Therefore,

$$\lVert f+g\rVert_2^2 \leq ( \lVert f\rVert_2+\lVert g\rVert_2)^2$$

Finally,

$$\lVert f+g\rVert_2 \leq \lVert f\rVert_2+\lVert g\rVert_2 $$

$\endgroup$
4
  • 1
    $\begingroup$ This is the correct way (regarding insights into the structure). Of course, this perhaps leaves the OP with the instructive task of showing that the dot product is indeed an inner product $\endgroup$
    – Hagen von Eitzen
    May 22, 2021 at 8:09
  • $\begingroup$ I'm just wondering what $( \cdot |\cdot )$ notation means? $\endgroup$
    – user899971
    May 24, 2021 at 9:12
  • 1
    $\begingroup$ It is the same thing as for the square root function, I will denote it by $\sqrt{\cdot} \,$. So $(\cdot | \cdot)$ is the inner product function, defined by $(f|g) = \int_s^t f(x)g(x) \, \mathrm{d}x $. So $(\cdot | \cdot)$ is just a explicit name to refer to your inner product. Is it clear enough? $\endgroup$
    – Axel
    May 24, 2021 at 9:28
  • $\begingroup$ I could have called it $\varphi$ (or any name) instead of $(\cdot | \cdot)$, it is just common notations. $\endgroup$
    – Axel
    May 24, 2021 at 9:39
0
$\begingroup$

By Cauchy–Schwarz inequality $$\left\vert \int\limits_s^t g(x) h(x)\mathrm{dx} \right\vert \leqslant \Vert g \Vert_2 \cdot \Vert h \Vert_2 ,$$ hence $$\Vert g + h\Vert_2^2 =\Vert g\Vert _2^2 +\Vert h \Vert_2^2 + 2\cdot (g \cdot h) = \\=\Vert g\Vert_2^2 + \Vert h \Vert_2^2 + 2\cdot\int\limits_s^t g(x) h(x)\mathrm{dx} \leqslant \\ \leqslant \Vert g\Vert_2^2 + \Vert h \Vert_2^2 + 2 \Vert g \Vert_2 \cdot \Vert h \Vert_2 = \big( \Vert g \Vert_2 + \Vert h \Vert_2 \big)^2.$$

$\endgroup$

You must log in to answer this question.