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Let a $f$ be a function on $[a,b]$ such that it's differentiable on $(a,b)$ such that $|f'(x)| \leq M$ for some real $M$. Prove that $f$ is also bounded.

My proof: Fix an arbitrary $x_0 \in (a,b)$ then for all $k \in (a,x_0)$, then by MVT, there exist $c \in (k,x_0)$ such that $$f'(c)(x_0-k)+ f(k) = f(x_0)$$ Now $$|M(x_0 - k)|+|f(k)| \geq |f(x_0)| \implies |M(x_0 - a)|+|f(k)| > |f(x_0)|$$ Hence, for some $L$, $L> |f(x)|$ for all $x$. $\blacksquare$

My concern about this solution is that, I'm not sure if my arguments like "fix any $x_0$" is correctly put, and hence if this solution works for all $x \in (a,b)$. I haven't done a lot of proofs, where it's required to fix some arbitrary point in domain, so would be much appreciated if someone checks if all the arguments are correct.

Gary
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zaemon_23
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1 Answers1

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Intuitively, as $|f'(x)| < M$ for all $x \in (a,b)$, the function $f(x)$ should always be below the (affine) linear function $g(x) = M(x-a) + f(a)$. So the idea is that $$f(x) \leq g(x) \quad \forall x \in [a,b]\tag{1}\label{eq:1}$$ should be true. Clearly, $g$ is bounded on $[a,b]$, because any linear function is bounded on a finite interval (it has a maximum at $x = b$ since $M>0$). So if we prove \eqref{eq:1}, the function $f$ will be bounded from above. To get bounded from below, make a similar argument with $h(x) = -M(x-a) + f(a)$.

Proof of \eqref{eq:1}. Suppose to the contrary that there is a point $x_0 \in (a,b)$, where $f(x_0) > g(x_0)$. Then $$f(x_0) - f(a) > g(x_0) - f(a) = g(x_0) - g(a) = M(x_0-a).\tag{2}\label{eq:2}$$ By the mean value theorem, there exists a point $x_1 \in (a,x_0)$ with $$f'(x_1) = \frac{f(x_0) - f(a)}{x_0 - a} > M,$$ where the inequality comes from \eqref{eq:2}. This constradicts the assumption that $f'(x) \leq M$ for all $x \in (a,b)$. $\square$

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