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If we take the Banach spaces $(X, \|\cdot \|_X )$ and $(Y, \|\cdot \|_Y )$ and the bounded linear map $F: X \rightarrow Y$, then the norm of $F$ is $$\|F\| = \underset{\|x\|_X = 1}{\textrm{sup}} \|F(x)\|_Y.$$ My question is how we can determine, generally, whether or not the map $F$ actually achieves that supremum. This is the example I have in mind: $(X, \|\cdot \|_X ) = (C([0, 1], \mathbb{R}), \|\cdot \|_{\infty})$ and $(Y, \|\cdot \|_Y)= (\mathbb{R}, |\cdot |)$, and the map $F: X \rightarrow Y$ is given by $F(f) = \int_{0}^{1} xf(x)dx$. It can be shown in a standard way that $\|F\| = \frac{1}{\sqrt3}$ and that $F$ actually achieves this norm when $f(x) = x$.

But now consider the situation where instead of $X$ being all continuous functions from $[0,1]$ to $\mathbb{R}$, it's only those functions where $f(1)=0$. I believe it can be shown that $F$ doesn't achieve its norm in this case, although we can show that the norm of $F$ remains $\frac{1}{\sqrt3}$ by taking the limit of $|F(f_n )|$ where $$f_n (x) = \begin{cases} x & 0\leq x\leq 1-\frac{1}{n} \\ (1-n)x + (n-1)& 1-\frac{1}{n}< x\leq 1 \end{cases} $$

and this limit will be $\frac{1}{\sqrt3}$ because $f_n \rightarrow x$ on $[0,1]$. But since $f(x) = x$ isn't in our new space, we can't use it to prove that $F$ reaches its norm here. But we don't necessarily know that there's only one function that lets $F$ achieve its norm, do we? Maybe a theorem of Riesz can be used to show this? Otherwise, how can we show that $F$ won't achieve its norm on this new space? And how can we do that generally?

$\bf{\textrm{EDIT}}$: I miscalculated the norm of $F$ -- it is actually $\frac{1}{2}$.

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  • $\begingroup$ Isn't $\int_0^1 x^2=\frac{1}{3}$? $\endgroup$ – Aweygan Mar 19 '18 at 14:12
  • $\begingroup$ In any case, assume that there is some $f$ with $f(1)=0$, $\|f\|=1$ and $F(f)=\|F\|$. Show that $|f(x)|<\frac{1}{2}$ in some neighborhood of $1$, This shows that $|xf(x)|<\frac{1}{2}|x|$ in some neighborhood of $1$, which should help you obtain a contradiction. $\endgroup$ – Aweygan Mar 19 '18 at 14:17
  • $\begingroup$ @Aweygan Yes, sorry about that. The function that gives us the norm is actually $f(x) = \frac{2}{\sqrt3}$. Still, this isn't in the changed space either. And thanks, let me see if I can use your suggestion. $\endgroup$ – SPS Mar 19 '18 at 14:21
  • $\begingroup$ Actually that choice of $f$ won't work since it doesn't have the right supremum... I'll have to think about which function will give us the norm we need. $\endgroup$ – SPS Mar 19 '18 at 14:27
  • $\begingroup$ OK, I see that I've over-complicated things by using the wrong inequalities in my working. The norm is actually $\frac{1}{2}$, not $\frac{1}{\sqrt3}$, and we can show that $F$ achieves this norm in the original case simply when $f(x) = 1$. The question still stands, though, how can we show in general that a linear map does or does not achieve its norm. $\endgroup$ – SPS Mar 19 '18 at 14:41
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Firstly, when considering $F$ as a map from $C([0,1])$ we have $\|F\|=\frac{1}{2}$. One way to see this is that $$\left|\int_0^1 xf(x)\ dx\right|\leq\int_0^1|x||f(x)|\ dx\leq\|f\|\int_0^1x\ dx=\frac{1}{2}\|f\|,$$ and the constant function $f(x)=1$ gives us this upper bound.

Now let's write $\tilde X=\{f\in C([0,1]):f(1)=0\}$ and talk about $F$ restricted to $\tilde X$. We still have $\|F\|=\frac{1}{2}$, since we can take approximations to the constant $1$ function while staying in $\tilde X$.

But to show that $F$ does not achieve this norm is a different matter. To answer your last question, there isn't a general way to do this, but your best bet is to go by contradiction since you at least get a function to play with. So assume $f\in\tilde X$ with $\|f\|=1$ and $F(f)=\frac{1}{2}$. Since $f(1)=0$, take any $\varepsilon>0$ and use continuity to show there is some $x_0\in (0,1)$ such that $|f(x)|<\varepsilon$ whenever $x_0<x\leq1$. Then we have $$\frac{1}{2}=|F(f)|\leq\int_0^1|xf(x)|\ dx=\int_0^{x_0}|xf(x)|\ dx+\int_{x_0}^1|xf(x)|\ dx<\frac{1}{2}x_0+\varepsilon(1-x_0).$$ If $\varepsilon<\frac{1}{2}$ (which is OK, since it was arbitrarily positive), we obtain a contradiction.

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As noted by @Aweygan, we have $\|F\| = \frac12$ on $C[0,1]$ and on $\tilde{X} = \{f \in C[0,1] : f(1) = 0\}$.

We will show that if $F$ achieves its norm for some $f \in C[0,1]$, then necessarily $f \equiv 1$ or $f \equiv -1$.

Assume $F(f) = \frac12$ for some $f \in C[0,1]$, $\|f\| = 1$.

Since $\|f\| = 1$ we have $f(x) \le 1, \forall x\in[0,1]$.

If $f(x_0) = 1 - \varepsilon < 1$ for some $x_0 \in [0,1]$ and $\varepsilon > 0$ then there exists an interval $I$ such that $x_0 \in I \subseteq [0,1]$ and $f(x) \le 1-\frac\varepsilon2, \forall x\in I$.

We have

\begin{align} F(f) &= \int_{[0,1]}xf(x)\,dx \\ &= \int_{I}x\underbrace{f(x)}_{\le 1-\frac\varepsilon2}\,dx + \int_{[0,1]\setminus I}x\underbrace{f(x)}_{\le 1}\,dx \\ &=\underbrace{\left(1-\frac\varepsilon2\right)}_{<1}\int_I x\,dx + \int_{[0,1]\setminus I} x\,dx \\ &< \int_{[0,1]}x\,dx \\ &= \frac12 \end{align}

which is a contradiction with $F(f) = \frac12$. Therefore $f \equiv 1$.

Similarly, if $F(f) = -\frac12$ then we get $f \equiv -1$.

Either way, the constant functions $1$ and $-1$ are not in $\tilde{X}$ so $f$ does not attain its norm on $\tilde{X}$.

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  • $\begingroup$ Thanks, that's also a nice argument, and nicely laid out. $\endgroup$ – SPS Mar 20 '18 at 13:12

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