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 $\frac{1}{2}$.

  • $\begingroup$ 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

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.


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 $\tilde{X}$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.