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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?

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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.

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