User talk:Sciyoshi

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= Problems =

1. Let $$A \subset \mathbb{R}$$, and suppose that $$\forall {x \in A}, \forall {n \in \mathbb{N}}, | f_n(x) | \leq M_n$$. If the sequence $$(f_n)$$ converges uniformly to $$f$$ on $$A$$, show that:

a) $$f$$ is bounded on $$A$$

= Solutions =

1.

a) Suppose not, that is

$$\forall {M \in \mathbb{R}}, \exists {x \in A} \mbox{ s.t. } |f(x)| > M$$.

Fix $$\epsilon_0 > 0$$ sufficiently small. Since the convergence is uniform,

$$\exists {K_{\epsilon_0} \in \mathbb{N}} \mbox{ s.t. } \forall {n \in \mathbb{N}, n \geq K_{\epsilon_0}}, \forall {x \in A}, | f_n(x) - f(x) | < \epsilon_0$$.

In particular,

$$\forall {x \in A}, | f_{K_{\epsilon_0}}(x) - f(x) | < \epsilon_0$$.

Now, by contradiction,

$$\exists {x_0 \in A} \mbox{ s.t.} |f(x_0)| > M_{K_{\epsilon_0}} + \epsilon_0$$.

But then, since by assumption

$$|f_{K_{\epsilon_0}}| < M_{K_{\epsilon_0}}$$,

$$|f_{K_{\epsilon_0}}(x_0) - f(x_0)| \geq | f(x) | - | f_{K_{\epsilon_0}}(x_0) | > M_{K_{\epsilon_0}} + \epsilon_0 - M_{K_{\epsilon_0}} = \epsilon_0$$,

contradiction.