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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?

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  • $\begingroup$ Why do you write $\|K(f) \|_{L(X,Y)}$? $K(f)$ is simply an element in $F$. $\endgroup$
    – Demophilus
    Commented Nov 2, 2017 at 19:50
  • $\begingroup$ Also, correct me if I'm wrong, but wouldn't $C = \sup_{s,t \in [0,1]} \lvert k(s,t) \rvert$? $\endgroup$
    – Demophilus
    Commented Nov 2, 2017 at 19:53
  • $\begingroup$ 1. isn't $||K(f)||_{L(E,F)}=C$? Why shouldn't I write that? 2. I am not sure. How did you come up with that solution? $\endgroup$
    – Tobi92sr
    Commented Nov 2, 2017 at 20:07
  • $\begingroup$ 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. $\endgroup$
    – Demophilus
    Commented Nov 2, 2017 at 20:12

1 Answer 1

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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 $\|K\|_{L(E,F)} \leq C$.

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  • $\begingroup$ 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... $\endgroup$ Commented Nov 2, 2017 at 20:50
  • $\begingroup$ @TheoreticalEconomist You're absolutely right, I was still thinking about that. $\endgroup$
    – Demophilus
    Commented Nov 2, 2017 at 20:58
  • $\begingroup$ 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 $C = \sup_{s \in [0,1]} \int_0^1 \lvert k(s,t) \rvert \, \mathrm d t$. $\endgroup$ Commented Nov 2, 2017 at 21:01
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    $\begingroup$ @TheoreticalEconomist He apparently edited it out for some reason, and as I've said I'm trying to prove that. $\endgroup$
    – Demophilus
    Commented Nov 2, 2017 at 21:09
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    $\begingroup$ @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 $\|K(f)\|_\infty > C \|f\|_1$. $\endgroup$
    – Demophilus
    Commented Nov 2, 2017 at 21:32

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