# Show that a differentiable function $f:\mathbb{R} \to \mathbb{R}$ has a global max in $a$ if $a$ is its local max

Let $f:\mathbb{R} \to \mathbb{R}$ be a differentiable function and assume that the only stationary point $f$ has is a local max in the point $A = (a,f(a))$. Show that $A$ must be a global max for $f$.

My work so far:

Since $f$ is differentiable we have that $\ f'(a) = 0$. Since $A$ is the only stationary point and local max for $f$, we get:

$$\forall x < a \implies f'(x) > 0 \ \land \forall y > a \implies f'(y) < 0.$$

It then follows that:

$$f(x),\ f(y) < f(a).$$

Which is what we wanted to show.

Is this good enough? Need someone to verify since there are no answers to this in the book I'm using (Multivariate calculus w/linalg by Tom Lindstrøm).

• what does $A(a,f(a))$ mean? – user7530 May 5 '16 at 7:19
• Did you want to write differentiable function rather than differential function? – Martin Sleziak May 5 '16 at 8:32
• I think that adding an exact reference rather than saying "my book" would also improve the question. (Not to mention that the phrase my book is somewhat ambiguous. It might mean a book your studying or a book you wrote.) – Martin Sleziak May 5 '16 at 8:34
• It seems to me that you are using the continuity of the derivative, and this is not assumed. – Siminore May 5 '16 at 9:13
• Sorry, A is the point, differentable edited, book added, and I'm just saying that since f is differentiable, then it's derivative must me positive to the left and negative to the right. if not, then the statements will not hold and a can't be a stationary point. – Thomas May 5 '16 at 11:01

If $f$ is constant on a small neighborhood of $a$, then $f$ has infinitely many critical points. Hence, recalling that $A$ is a local maximum point, there exists $\delta>0$ such that $$f(a-\delta) < f(a), \quad f(a)> f(a+\delta).$$ Assume that there exists a point $b$ such that $f(b)>f(a)$. Without loss of generality, we may suppose that $b>a+\delta>a$. Then $\min_{a \leq x \leq b} f(x)$ is attained inside $(a,b)$, and by Fermat's theorem there exists anohter critical point. This is a contradiction.