Here is the Gelfand lemma:Suppose $X$ is a Banach space, $p:X\to\mathbb{R}$ satisfies :
- $p(x)\ge 0$;
- $p(\lambda x)=\lambda p(x),\quad \forall \lambda>0$;
- $p(x_1+x_2)\le p(x_1)+p(x_2)$;
- $\liminf\limits_{n\to\infty}p(x_n)\ge p(x)$ where $x_n\to x$.
Then there exists $M>0$ such that $p(x)\le M||x||$.
I have proved it and try to use it to prove the uniform boundness principle:
Let $X,Y$ be Banach spaces and $\{A_\lambda:\lambda\in\Lambda\}$ be a family of bounded linear operators from $X$ to $Y$. If $\{A_\lambda:\lambda\in\Lambda\}$ is pointwise bounded on X, then $\{A_\lambda:\lambda\in\Lambda\}$ is uniformly bounded, i.e., $\sup∥T_λ∥<\infty$.
I considered to show that $\sup\limits_{\lambda\in\Lambda}||A_\lambda x||$ is the $p(x)$ in Gelfand Lemma but I failed.
Your initial attempt at defining $p: X \to \mathbb{R}$ by $p(x) := \sup_{\lambda \in \Lambda} \| A_{\lambda} (x) \|$ is indeed a good approach. The most challenging part is to show that the fourth result holds.
To show this, let $(x_{n})_{n \in \mathbb{N}}$ be a sequence in $X$ with limit $x$. Let $\lambda \in \Lambda$. By the continuity of $A_{\lambda}$ and the norm, it follows that $\lim_{n \to \infty} \| A_{\lambda}(x_{n}) \| = \|A_{\lambda}(x)\|$. In particular, $\|A_{\lambda}(x)\| \leq \liminf_{n\to\infty} \|A_{\lambda}(x_{n})\|$. As a direct result of this,
$$ \| A_{\lambda}(x) \| \leq \liminf_{n \to \infty} \|A_{\lambda}(x_{n})\| \leq \liminf_{n \to \infty} \left( \sup_{\gamma \in \Lambda} \|A_{\gamma}(x_{n}) \| \right) = \liminf_{n \to \infty}p(x_{n}).$$
As this holds for all $\lambda \in \Lambda$, $\liminf_{n \to \infty}p(x_{n})$ is an upper bound for $\{\|A_{\lambda}(x)\| : \lambda \in \Lambda \}$. But as $p(x) = \sup \{\|A_{\lambda}(x)\| : \lambda \in \Lambda \}$, it follows that $p(x) \leq \liminf_{n \to \infty}p(x_{n})$.
The other three parts should be straightforward to verify from the definitions.