Proving a norm of a set of continuous functions (Specifically the triangle inequality part)

Question

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

Answer 1

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,

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

Answer 2

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 $$\Vert g + h\Vert_2^2 =\Vert g\Vert _2^2 +\Vert h \Vert_2^2 + 2\cdot (g \cdot h) = \\=\Vert g\Vert_2^2 + \Vert h \Vert_2^2 + 2\cdot\int\limits_s^t g(x) h(x)\mathrm{dx} \leqslant \\ \leqslant \Vert g\Vert_2^2 + \Vert h \Vert_2^2 + 2 \Vert g \Vert_2 \cdot \Vert h \Vert_2 = \big( \Vert g \Vert_2 + \Vert h \Vert_2 \big)^2.$$