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I am trying to prove the following statement

For $1

$\|S(t) w \|_{L^p} \leq t^{\frac{1}{p}-\frac{1}{q}} \|w\|_{L^q}$ for all $t>0$

and for each $w \in L^q(\Omega)$

where $S(t)$ is an analityc semigroup and $\Omega \subset R^{n}$ is open and bounded.

So far I have proved my claim in this cases:

i) $1

ii) $1

So to finish the proof I only lack the case $p=q=\infty$

I want to know if my idea to finish the test is correct or else can you suggest me how to complete the demonstration?

My idea is this, proof $p=q=\infty$, Let $w \in L^{\infty}(\Omega)$ then $w \in L^q$ for all $q<\infty$ and apply case i) we get

$\|S(t) w \|_{L^q} \leq \|w\|_{L^q}$

Then taking the limit when p tends to infinity, we arrive to

$\|S(t) w \|_{L^{\infty}} \leq \|w\|_{L^{\infty}}$ this is correct? I assume that $\lim_{p \to \infty} \|w\|_p=\|w\|_{\infty}$

Thanks for your help and your comments I have been improving the post and sorry for my english.

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    What is $t$? I assume you mean that $T$ is a bounded linear operator. – 2017-02-03
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    Yes $T$ is bounded linear operator, but i think that this is not essential to complete the proof, but thanks for your comment, i edited the post. – 2017-02-03
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    I still don't see where lower-case $t$ is defined. – 2017-02-03
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    i´m sorry, $t > 0$ is a real number. – 2017-02-03
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    I want to describe the problem in an abstract way. This statement is not true for all bounded linear operators, Really $T$ is very specific – 2017-02-03
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    If $p \ne q$, the statement: "$\forall t > 0 : \|Tw\|_{L^p} \le t^{1/q-1/p} \|w\|_{L^q}$" is equivalent to "$\|Tw\| \le 0$ by considering $t \to 0$ or $t \to \infty$. There should be another place in this question where $t$ comes into play. – 2017-02-03
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    Thanks, I must be more precise, the operator $T$ depends on t, in fact $T=S(t)$ where $S$ is an analytical semigroup. sorry – 2017-02-03

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