Finding supremum of a weird function

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Finding supremum of a weird function

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pokoknyaakuimut

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pokoknyaakuimut

#1 h Feb 14, 2025, 11:10 AM

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Find $\text{sup}\{2^{2x}+2^{\frac{1}{2x}}:x\in\mathbb{R}, x<0\}$ . Easy to guess that the answer is $1$ , but I haven't found the reason yet.

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#2 h Feb 14, 2025, 1:39 PM • 2 Y

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Let $c = \ln 2$ and $t = - 2x$ , so we are looking for the supremum of $f(t) = e^{-ct} + e^{-\frac c t}$ for $t > 0$ . By the $t \leftrightarrow \frac 1 t$ symmetry it sufficres to consider $t \ge 1$ . Note that $f(1) = 2 e^{-c} = 1$ and $f(t) \to 1$ for $t \to \infty$ , also it is easy to see that $f(2)< 1$ , so it takes a global minimum at some point in $(1,\infty)$ (by continuity and boundary values) and it suffices to show that it does not take a global maximum in $(1,\infty)$ . (This implies that the global maximum is attained at $t = 1$ ).

Let us now look for possible local extrema of $f$ on $(1,\infty)$ . We have $f'(t) = -ce^{-ct} + \frac c {t^2} e^{- \frac c t}$ , so
$f'(t) = 0 \Leftrightarrow t^2 = e^{c(t - \frac 1 t)} \Leftrightarrow g(t) := c (t - \frac 1 t) - 2 \ln(t) = 0.$ We note that $g'(t) = c(1 + \frac 1 {t^2}) - \frac 2 t$ and $g''(t) = - \frac {2c} {t^3} + \frac 2 {t^2}> 0$ for $t \ge 1$ (sinc $c = \ln(2) < 1 y\le t$ ). So $g$ is strictly convex and has a zero in $t = 1$ , and so it can have at most one further zero in $(1,\infty)$ .

Thus $f$ can have at most one local max/min in $(1,\infty)$ , and by the above we are done.

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removablesingularity

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removablesingularity

#3 h Mar 25, 2025, 11:32 PM

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name $y = -x>0, 2^{2x} + 2^{\frac{1}{2x}} = \frac{1}{2^{2y}} + \frac{1}{2^{\frac{1}{2y}}}.$ Take $A = \frac{1}{2},2^{2x} + 2^{\frac{1}{2x}} = \frac{1}{2^{2y}} + \frac{1}{2^{\frac{1}{2y}}} = A^{2y} +\frac{1}{A^{\frac{1}{2y}}}\ge 2\sqrt{A^{2y}.\frac{1}{A^{\frac{1}{2y}}}}=2\sqrt{A^{2y+\frac{1}{2y}}}\ge 2\sqrt{A^{2\sqrt{2y.\frac{1}{2y}}}}=2A=1$ .

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#4 h Mar 26, 2025, 6:32 AM

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@above: this does not work. Note that $A < 1$ , so the last inequality is reversed.

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MihaiT

#5 h Mar 26, 2025, 6:56 AM

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solyaris wrote:

Let $c = \ln 2$ and $t = - 2x$ , so we are looking for the supremum of $f(t) = e^{-ct} + e^{-\frac c t}$ for $t > 0$ . By the $t \leftrightarrow \frac 1 t$ symmetry it sufficres to consider $t \ge 1$ . Note that $f(1) = 2 e^{-c} = 1$ and $f(t) \to 1$ for $t \to \infty$ , also it is easy to see that $f(2)< 1$ , so it takes a global minimum at some point in $(1,\infty)$ (by continuity and boundary values) and it suffices to show that it does not take a global maximum in $(1,\infty)$ . (This implies that the global maximum is attained at $t = 1$ ).

Let us now look for possible local extrema of $f$ on $(1,\infty)$ . We have $f'(t) = -ce^{-ct} + \frac c {t^2} e^{- \frac c t}$ , so
$f'(t) = 0 \Leftrightarrow t^2 = e^{c(t - \frac 1 t)} \Leftrightarrow g(t) := c (t - \frac 1 t) - 2 \ln(t) = 0.$ We note that $g'(t) = c(1 + \frac 1 {t^2}) - \frac 2 t$ and $g''(t) = - \frac {2c} {t^3} + \frac 2 {t^2}> 0$ for $t \ge 1$ (sinc $c = \ln(2) < 1 y\le t$ ). So $g$ is strictly convex and has a zero in $t = 1$ , and so it can have at most one further zero in $(1,\infty)$ .

Thus $f$ can have at most one local max/min in $(1,\infty)$ , and by the above we are done.

very nice!

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