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I am looking for a proof of the following statement which is known to be true as far as I heard.

Let $g\colon [a,b]\to \mathbb{R}$ be a smooth function. Assume that $$b-a\leq \pi.$$$$b-a< \pi.$$ Assume also $$g(a)\geq 0,g(b)\geq 0,$$ $$g''+g\leq 0 \mbox{ on } [a,b].$$ Then $g\geq 0$ on $[a,b]$.

I am looking for a proof of the following statement which is known to be true as far as I heard.

Let $g\colon [a,b]\to \mathbb{R}$ be a smooth function. Assume that $$b-a\leq \pi.$$ Assume also $$g(a)\geq 0,g(b)\geq 0,$$ $$g''+g\leq 0 \mbox{ on } [a,b].$$ Then $g\geq 0$ on $[a,b]$.

I am looking for a proof of the following statement which is known to be true as far as I heard.

Let $g\colon [a,b]\to \mathbb{R}$ be a smooth function. Assume that $$b-a< \pi.$$ Assume also $$g(a)\geq 0,g(b)\geq 0,$$ $$g''+g\leq 0 \mbox{ on } [a,b].$$ Then $g\geq 0$ on $[a,b]$.

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asv
asv
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  • 54
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Elementary inequality generalizing convexity of a function on a segment

I am looking for a proof of the following statement which is known to be true as far as I heard.

Let $g\colon [a,b]\to \mathbb{R}$ be a smooth function. Assume that $$b-a\leq \pi.$$ Assume also $$g(a)\geq 0,g(b)\geq 0,$$ $$g''+g\leq 0 \mbox{ on } [a,b].$$ Then $g\geq 0$ on $[a,b]$.