Proof that Integral is continuous in given metric space

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I am supposed to show that the following statement is true:

For $p \in [1, \infty]$ the projection of $I:(C([a,b]), \|\cdot\|_p) > \rightarrow (\mathbb{R},|\cdot|)$ defined as

$$I(f) = \int_{a}^{b}f(x) dx,$$ is continous.

I tried proving this statement with the $\epsilon$-$\delta$-criteria, since this was a hint I was given, sadly I couldn't really find a way to define my $\epsilon$ and my $\delta$ so I tried it with another approach.

In my script I found a very handy Satz, though. Which says:

Satz 4.29: Let $(X, \|\cdot\|_X)$ and $(X', \|\cdot\|_{X'})$ be normed vector spaces over $\mathbb{K}$, where $A:X\rightarrow X'$ is linear, then the following statements are equivalent:

  • A is continous
  • A is continous at 0
  • There exists a $C > 0$ with $\|A(x)\|_{X'} \leq C\|x\|_X$ for every $x \in X$

So I tried proofing the upper statement with the Satz 4.29.

First of all I can say that $I(f)$ is linear since we already proved that Integrals are linear.

Then I claimed that $$\forall p \neq \infty:||f||_{p} \leq ||f||_{\infty}$$ Which is easy provable since the supremum-norm is just defined as the supremum of every function, thus being always greater or equal to any given $p$-norm.

With that I finally applied Satz 4.29:

$$|I(f)|\leq\int_{a}^{b}|f(x)|dx \leq (b-a)||f||_{\infty} \quad with \quad \forall p \neq \infty:||f||_{p} \leq ||f||_{\infty}$$

My questions:

  • Am I already done with the proof that way?
  • How is this solvable with the $\epsilon$-$\delta$-criteria? I think that this is a way smoother way to solve it, but I couldn't find a way

Thanks in advance.

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You can show that for each $p$ there exists a constant $C_p>0$ such that $|I(f)|\le C_p\|f\|_p$ for all your $f$'s, as you state in Satz 4.29. You showed it for $p=\infty$. You still need to show it for $p\neq\infty$. You can apply Hölder's inequality to $\int_a^b1|f(x)|dx$. You can apply Hölder's inequality to a $\epsilon$-$\delta$ argument bypassing Satz 4.29 (if you know that $|\int f-\int g|\le \int|f-g|$).