I am trying to prove:
Suppose $f$ is bounded and continuous on $[0, 1]$. For any $n \in \mathbb{Z}^+$, define $$||f||_n = \left(\int_0^1|f(x)|^n \, dx\, \right)^{1/n}$$ Then, $\lim_{n \to \infty} ||f||_n \ge \max\{|f(x)|:x \in [0, 1]\}$.
I found the following proof:
Let $\epsilon>0$ and $\max\{|f(x)|:x \in [0, 1]\}$. Then, $M-\epsilon<M$ which means that there exits a $a \in [0, 1]$ such that $|f(a)| \ge M-\epsilon$. Since $f$ is uniformaly continuous on $[0, 1], \exists \delta>0$ such that for all $x \in (a-\delta, a+\delta)$, we have $|f(x)|> M-\epsilon$. Now, let $\delta_1=\{\delta, 1/2\}$ which implies that $2\delta_1 \ge 1$. Hence,
$$\int_0^1|f(x)|^n \, dx = \int_0^{a-\delta_1}|f(x)|^n \, dx+ \int_{a-\delta_1}^{a+\delta_1}|f(x)|^n \, dx+\int_{a+\delta_1}^1|f(x)|^n \, dx \ge \int_{a-\delta_1}^{a+\delta_1}|f(x)|^n \, dx\ge \int_{a-\delta_1}^{a+\delta_1}(M-\epsilon)^n\, dx$$
Thus,
$$\int_0^1|f(x)|^n \, dx \ge |M-\epsilon|^n \cdot 2 \delta_1 \ge M-\epsilon^n$$
and the rest of the proof is rather trivial. Now, I am highly skeptical of this proof. For instance, the above proof seems to erroneously assume that $M-\epsilon \ge 0$. Also, how does the definition of uniform continuity imply that $|f(x)|> M-\epsilon$?
Is there a way to fix this proof? Can someone please suggest how this can be proven (without using any concepts from measure theory)? If someone can show how $|f(x)| \ge M$, I can complete the rest of the proof from there.
