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Are every analytic semigroups norm continuous? Are there counterexamples otherwise? What would make analytic semigroup norm continuous as well? (My apology, frankly, I am not sure if norm continuous is equivalent of saying uniform continuous for the semigroups.)

I know the difference between $C_0$ semigroup and norm continuous semigroup is that for norm continuous semigroup we have $$\lim_{t\to 0}\|S(t)-Id\|=0$$ in contrast to $$\lim_{t\to 0}S(t)x=x$$ for $C_0$ semigroup.

I saw from some text and have a feeling that every analytic semigroup is also norm continuous semigroup. (Background: This has a huge impact inspecting on different notions of solutions. For a norm continuous semigroup, it is known that weak, mild and strong solution are equivalent.)

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    What do you mean by uniformly continuous? Do you mean continuous in the operator norm? Only bounded generators have that property. The Poisson semigroup generated from the Poisson integral on the upper half plane is an analytic semigroup to play with. – 2017-01-04
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    Norm continuous I meant – 2017-01-04
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    What norm? Operator norm? Vector norm? Uniform with respect to what? – 2017-01-04
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    in the edited sense – 2017-01-04
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    An analytic semigroup can be norm continuous everywhere except at $t=0$. This has to do with the semigroup $S(t)$ for $t > 0$ being in the domain of every power of the generator. For example, with the heat semigroup, for any $t > 0$, functions are infinitely differentiable. The same is true of the Poisson semigroup $P(y)f = \frac{1}{\pi}\int_{-\infty}^{\infty}\frac{y}{(t-x)^2+y^2}f(t)dt$. A holomorphic semigroup is holomorphic in $t$ within a cone in the complex plane. The tip of the cone is where all the action is. – 2017-01-04
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    To further clarify, $S(t)f$ is differentiable in $t$ iff $f$ is in the domain of the generator $A$. Because $S$ is holomorphic for $t > 0$, then $S(t)f$ is in the domain of every power of the generator. Powers work nicely, too, because $(S(t)-I)f \sim t^{\alpha}f$ is directly related to $f$ being in the domain of the fractional power of the generator. – 2017-01-04

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From the Theorem below, you are wrong, since the generator of analytic semigroup is not necessarily bounded. enter image description here

Note that $T(t)$ is called uniformly continuous if and only if $\lim_{t\to 0^+}\|T(t)-I\|=0$. enter image description here