A problem in the space $C[a,b]$

Question

Let $E=C[a,b]$ provide with the $\max$ norm. Let $S\neq \emptyset$, let and $D(t,\lambda)$ be a continuous function (for each $\lambda\in S$), from $[a,b]$ to $\mathbb{R}$, such that $\displaystyle D=\sup_{\lambda}\int_{a}^b |D(t,\lambda)|dt<\infty$. We define for $f\in E$ $$A(f)=y(\lambda)=\int_{a}^b D(t,\lambda)f(t)dt$$ I have to prove that: $\|A\|=D$, where $\sup_{\|x\|=1}{\|A(x)\|}=\|A\|$.

My Approach: I have already showed that $A$ is a linear continuous operator, from $E$ to $F=\{f:S\to \mathbb{R} \mid \text{$f$ is bounded} \}$. Furthermore: since $$\|A(f)\|=\left\|\int_{a}^bD(t,\lambda)f(t)dt \right\|\leq \int_{a}^b |D(t,\lambda)|\cdot |f(t)|dt\leq \|f\| \int_{a}^b |D(t,\lambda)|dt$$ $$\leq \|f\|\cdot \sup_{\lambda}\int_{a}^b |D(t,\lambda)|dt=D\cdot \|f\|$$ we have $\|A\|\leq D$.

The difficult part is (I think) $\|A\|\geq D$. I have read some hint:
Consider

$f_n(t)=-1$ if $t<-1/n$,

$f_n(t)=1$ if $t>1/n$ and

$f_n(t)$ linear if $t$ is in $[-1/n,1/n]$.

How this can help me? Please any solution or explanation will be fully appreciated. Thanks!

Have you seen the definition of the norm of an operator? If so, you have already proved that $|A| \leq D$ right? — Prahlad Vaidyanathan, Jul 02 '14 at 14:32
What i know is not much, just that $||A||=\min{c\geq 0:||Av||\leq c||v||}$ for all $v$. — Valent, Jul 02 '14 at 14:40
@LuisValerin You showed above that $D$ is such a $c$, i. e. $|Af| \le D|f|$. Hence (as $|A|$ is the minimal $c$) you showed $|A| \le D$. — martini, Jul 02 '14 at 14:49
And for the second part? what i see in the hint is that all this functions $f_n$ have norm equal to 1. True right? — Valent, Jul 02 '14 at 14:58
Is $D(t,\lambda)$ real-valued or complex valued? [Real-valued will be simpler.] — Daniel Fischer, Jun 21 '15 at 20:00
According to your definition, $A$ depends not only on $f$ but also on $\lambda$, so what exactly is $||A||$? — sranthrop, Jun 21 '15 at 20:01

score 1 · Accepted Answer · answered Jun 22 '15 at 13:37

Since we already know that $\lVert A\rVert \leqslant D$, it remains to see that for every $\varepsilon > 0$ we have $\lVert A\rVert \geqslant D - \varepsilon$. If $D = 0$, that is clear, so let's assume $D > 0$.

Fix an arbitrary $0 < \varepsilon < D$. By definition of $\sup$, there is a $\lambda_0 \in S$ with

$$\int_a^b \lvert D(t,\lambda_0)\rvert\,dt > D-\frac{\varepsilon}{2}.$$

For ease of notation, let $g(t) = D(t,\lambda_0)$. We want to find a continuous $f$ with $\lVert f\rVert \leqslant 1$ so that

$$\int_a^b g(t)\cdot f(t)\,dt > D - \varepsilon.$$

For then we know

$$\lVert A\rVert \geqslant \lVert Af\rVert \geqslant \lvert (Af)(\lambda_0)\rvert = \biggl\lvert \int_a^b g(t)f(t)\,dt\biggr\rvert > D - \varepsilon,$$

and since $\varepsilon$ was arbitrary, it then follows that $\lVert A\rVert \geqslant D$.

This is the place where the $f_n$ come into play. If we did not require $f$ to be continuous, we could just take

$$\sigma(t) = \begin{cases} 1 &, g(t) > 0 \\ 0 &, g(t) = 0 \\ -1 &, g(t) < 0\end{cases}$$

and have $\lVert\sigma\rVert \leqslant 1$ and

$$\int_a^b g(t)\sigma(t)\,dt = \int_a^b \lvert g(t)\rvert\,dt > D - \frac{\varepsilon}{2}.$$

But generally $g$ is positive on some subintervals of $[a,b]$ and negative on others, so $\sigma$ is then not a continuous function. The idea is to approximate $\sigma$ by continuous functions. We set $\sigma_n(t) = 1$ when $g(t) > 1/n$, and $\sigma_n(t) = -1$ when $g(t) < -1/n$, and then we need to continuously extend $\sigma_n$ to the whole interval $[a,b]$ by defining it appropriately on the set $\{ t\in [a,b] : -1/n \leqslant g(t) \leqslant 1/n\}$.

Well, we simply take $\sigma_n(t) = f_n(g(t))$. Since $\sigma_n = f_n \circ g$, it is continuous. Also, by construction $-1 \leqslant \sigma_n(t) \leqslant 1$, so $\lVert \sigma_n\rVert \leqslant 1$. Further we have

$$\lim_{n\to\infty} \sigma_n(t) = \sigma(t)$$

for all $t\in [a,b]$, so we have good reason to expect

$$\lim_{n\to\infty} \int_a^b g(t)\sigma_n(t)\,dt = \int_a^b g(t)\sigma(t)\,dt = \int_a^b \lvert g(t)\rvert\,dt.\tag{1}$$

That is indeed the case. If you have some Lebesgue theory at your service, it follows immediately from the dominated convergence theorem. If not, that requires a little more work: Consider the function

$$\psi_n(t) = \lvert g(t)\rvert - g(t)\sigma_n(t).$$

Since $g$ and $\sigma_n$ are continuous, $\psi_n$ is continuous. If $\lvert g(t)\rvert \geqslant 1/n$, then we have $g(t)\sigma_n(t) = \lvert g(t)\rvert$ and hence $\psi_n(t) = 0$, and if $\lvert g(t)\rvert < 1/n$, then we have

$$\lvert \psi_n(t)\rvert \leqslant \lvert g(t)\rvert + \lvert g(t)\rvert\cdot \lvert \sigma_n(t)\rvert \leqslant 2\lvert g(t)\rvert \leqslant \frac{2}{n},$$

therefore

$$\biggl\lvert \int_a^b \lvert g(t)\rvert - g(t)\sigma_n(t)\,dt\biggr\rvert \leqslant \int_a^b \lvert \psi_n(t)\rvert\,dt \leqslant \frac{2}{n}(b-a),$$

which shows $(1)$.

From $(1)$, it follows that for all large enough $n$ we have

$$\int_a^b g(t)\sigma_n(t)\,dt > \int_a^b \lvert g(t)\rvert\,dt - \frac{\varepsilon}{2},$$

hence

$$\int_a^b g(t)\sigma_n(t)\,dt > D - \varepsilon$$

and $\lVert A\rVert > D-\varepsilon$, as desired.

Very elegant! And completetly understandable! +1 – Valent Jun 22 '15 at 14:30 — Valent, Jun 22 '15 at 14:30

A problem in the space $C[a,b]$

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