# Showing a Function is a Norm

Question: Fix an interval $$[a, b]$$. Let $$C[a,b]$$ be the set of continuous functions from $$[a,b]$$ to $$\mathbb{R}$$. For $$f,g \in C[a,b]$$, define a dot product and norm by $$f \cdot g = \int^b_a f(x)g(x) dx$$ and $$||f||_2 = \sqrt{f \cdot f}= \sqrt{\int^b_a |f(x)|^2 dx}$$ (note that the absolute value is not actually necessary). The dot product is clearly bilinear and symmetric (you do not need to show this or that $$\cdot$$ defines a dot product). Show that $$||.||_2$$ is a norm on $$C[a,b]$$.

Attempt:

I know that I need to show three things:

• $$||f||_2 \ge 0$$, with $$||f||_2=0$$ if and only if $$f=0$$ [Property A]
• $$||cf||_2 = |c|||f||_2$$ for all $$c \in \mathbb{R}$$ and $$f \in C[a,b]$$ [Property B]
• Triangle inequality: $$||f+g||_2 \le ||f||_2+ ||g||_2$$ for all $$f,g \in C[a,b]$$ [Property C]

For Property A: I am able to show that $$\int^b_a |f(x)|^2 dx$$ is nonnegative, but then get stuck. Since when you take the square root, $$\sqrt{\int^b_a |f(x)|^2 dx}$$, won't there be both a positive and negative solution? Also, to show $$||f||_2=0$$ if and only if $$f=0$$, I need to show both directions ($$a$$ iff $$b$$ is equivalent to $$a$$ implies $$b$$ AND $$b$$ implies $$a$$). I have no issue showing $$f(x)=0$$ implies $$||f||_2 = 0$$ but get stuck showing the other way around. I get to $$0 = \int^b_a f(x)^2 dx$$ but am not sure where to go from here (I'm guessing I must implement the fundamental theorem of calculus?).

I had no issue showing Property B holds as it is a simple matter of taking a constant out of an integral and then taking the square root.

For Property C I get really stuck. I need to show $$\sqrt{\int^b_a (f+g)^2 dx} \le \sqrt{\int^b_a f^2 dx} + \sqrt{\int^b_a g^2 dx}$$. My lecturer mentioned using the Cauchy–Schwarz inequality but I don't see it being of any value since it gives $$|x \cdot y| \le ||x|| ||y||=\sqrt{x\cdot x} \sqrt{y \cdot y}$$ which is a product rather than a sum.

Does anyone have any pointers?

For part (a), first let me clarify that if $$a$$ is a non-negative real, then $$\sqrt{a}$$ denotes the unique non-negative real whose square is $$a$$.

Now if $$f:[a, b]\to \mathbb R$$ is a continuous function and $$\int_a^bf^2\ dx = 0$$, then $$f$$ must be identically zero. This is becasue if $$f(x_0)=u\neq 0$$ for some $$x_0\in [a, b]$$, then in a small enough neighborhood $$(x_0-\delta, x_0+\delta)\cap[a, b]$$ around $$x$$, we would have $$f$$ takes values in $$[u/2, \infty)$$. This is becasue of the continuity of $$f$$. Thus $$\int_a^b f^2\ dx \geq \delta u^2/4>0$$ and we would have a contradiction.

For the Cauchy-Schwarz, you'd have to first construct a relevant vector space with an inner product. Here is how to do it.

Write $$V=C[a, b]$$ and note that $$V$$ is naturally a vector spaceunder pointwise addition and scaling by reals.

Now define an inner product on $$V$$ by writing $$\langle f, g\rangle = \int_{a}^b fg\ dx$$. It is easy to check that this is indeed an inner product.

Cauchy-Schwarz now becomes

$$\int_a^b fg\ dx \leq \sqrt{\int_a^b f^2\ dx}\sqrt{\int_a^b g^2\ dx}$$

If you sqaure both the sides of

$$\sqrt{\int^b_a (f+g)^2 dx} \le \sqrt{\int^b_a f^2 dx} + \sqrt{\int^b_a g^2 dx}$$ you will see that you are just left with the avatar of Cauchy-Schwarz.

You might want to look at $$L^2$$-spaces for more context.