Operatornorm inequality

Question

Let $$E:=(C^0([0,1]),\|\cdot\|_1) , \quad F:=(C^0([0,1]),\|\cdot\|_\infty),$$ with $$\|f\|_1:=\int_0^1\vert f(t)\vert \, dt,\quad \|f\|_\infty:=\sup_{t\in [0,1]}\vert f(t)\vert .$$ For $k \in C^0([0,1]\times[0,1])$ and $f\in E$ define $$K(f):=s\mapsto \int_0^1k(s,t)f(t)\,dt, s \in[0,1]$$ Show that $K$ is a continuous linear transformation $E\to F$ and find the operatornorm $C\ge0$, so $x\in E$, so that $\|K(f)\|_\infty\le C\|f\|_1$

I already showed the linearity. So I can use a theorem: Let $X$ and $Y$ be normed spaces and $f:X\to Y$ is linear, then the following are equivalent:

i) $f$ is continuous

ii) $\|f\|_{L(X,Y)}:=\sup_{\|x\|_X{\le1}}\|T_X\|_Y<\infty$

In this context is $\|f\|_{L(X,Y)}$ the Operatornorm $C$. I tried to do the following:

I suppose $\|f\|_1=1$ should be one condition for $C$ in order to be the supremum in $$\|K(f)\|_{L(X,Y)}:=s\rightarrow \left|\int_0^1k(s,t)f(t)\,dt\right|, \quad s\in [0,1]$$ $$\le s\rightarrow\int_0^1|k(s,t)f(t)|\,dt,\quad s\in[0,1]$$ $$\le s\rightarrow\int_0^1|k(s,t)||f(t)|\,dt,\quad s\in [0,1]$$ My question is: Is there a way to get rid of the $f(t)$ in the equation? I guess the final C should be $$C=\sup_{s\in[0,1]}\int_0^1|k(s,t)|\,dt$$ But I don't know how to do the final step(s). Can someone help me?

Why do you write $|K(f) |_{L(X,Y)}$? $K(f)$ is simply an element in $F$. — Demophilus, Nov 02 '17 at 19:50
Also, correct me if I'm wrong, but wouldn't $C = \sup_{s,t \in [0,1]} \lvert k(s,t) \rvert$? — Demophilus, Nov 02 '17 at 19:53
No $K(f)$ refers to an element of the vector space $F$, it isn't an operator. $K$ is the operator. That is, you're looking for $|K|{L(E,F)} = C$. And it just a guess, you somehow need to find $C$ such that the inequality $|K(f)|\infty \leq C |f|_1$ holds. But if we take you guess, the right hand side of that inequality would contain the product of two integrals while the left hand side contains only one integral. I don't see how that could work, which is probably why you're stuck. — Demophilus, Nov 02 '17 at 20:12

score 2 · Accepted Answer · answered Nov 02 '17 at 20:42

2

Let $C = \sup \limits_{s,t \in [0,1]} \lvert k(s,t)\rvert$. Note that for any $f \in E$ we have $$ \|K(f) \|_\infty = \sup_{s \in [0,1]} \big\lvert \int_0^1 k(s,t) f(t) dt \big\rvert. $$ For any $s \in [0,1]$ we have $$ \big\lvert \int_0^1 k(s,t) f(t) dt \big\rvert \leq \int_0^1 \lvert k(s,t) f(t) \rvert dt \leq \int_0^1 C \lvert f(t)\rvert dt= C \|f\|_1. $$ So we have $\|K\|_{L(E,F)} \leq C$.

answered Nov 02 '17 at 20:42

Demophilus

3,982

Don’t you need to prove that $\sup_{s,t\in [0,1]} \lvert k(s,t) \rvert$ is the smallest such $C$? The OP did ask for the operator norm... – Theoretical Economist Nov 02 '17 at 20:50
@TheoreticalEconomist You're absolutely right, I was still thinking about that. – Demophilus Nov 02 '17 at 20:58
I remember seeing this questions yesterday, and thought of answering it in exactly the way you just did. However, I could not prove that your $C$ is the operator norm. I think the OP deleted it and posted this new one. Someone there proposed $C = \sup_{s \in [0,1]} \int_0^1 \lvert k(s,t) \rvert , \mathrm d t$. – Theoretical Economist Nov 02 '17 at 21:01
@TheoreticalEconomist That's also what the OP proposed this time. But as I explained in the comments, I doubt that could be the case. – Demophilus Nov 02 '17 at 21:05
I can’t see where the OP proposed that... in any case, you should still prove that the $C$ you’ve defined is the operator norm. – Theoretical Economist Nov 02 '17 at 21:08
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@TheoreticalEconomist He apparently edited it out for some reason, and as I've said I'm trying to prove that. – Demophilus Nov 02 '17 at 21:09
Great, looking forward to it (genuinely). – Theoretical Economist Nov 02 '17 at 21:10
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@TheoreticalEconomist Haven't been able to prove it yet, but I do think I've found a concrete counter example as to why $C = \sup_{s \in [0,1]} \int_0^1 \lvert k(s,t) \rvert dt$ can't be the operator norm. That is take $k(s,t) = (t-t^2)$ and $f(t) = t-t^2$. Then $C = |f|1 = \frac{1}{6}$ but $|K(f)|\infty = \frac{1}{30}$ so $|K(f)|_\infty > C |f|_1$. – Demophilus Nov 02 '17 at 21:32
Yes, I believed you before you furnished the counter-example. An approach that I think may work is to construct a sequence ${ f_n } \subset E$ with $| f_n | = 1$ such that $| K(f_n) |_\infty \to C.$ Of course, one would need to appeal to an equivalent definition of the operator norm, but I think this is doable. – Theoretical Economist Nov 02 '17 at 21:57

Operatornorm inequality

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