Estimating a function from its derivative

Question

While modeling a heat transfer problem I came across the following estimation problem:

Let $ f: [0,\, 1] \to \mathbb R^+ $ and $g:[0,\, 1] \to \mathbb R^+$, which satisfy $f(0) = g(0)$ and $f''(x) < g''(x)$ for all $x \in [0,\, 1]$.

Now, does it hold true that $f(x) < g(x)$ for all $x \in [0,\, 1]$ as well?

As a start I used Taylor expansions for $f$ and $g$, but I can not find an estimate for the first order terms. If the statement does not hold true, which additional constraint would be required to make it true?

Thank you for your time.

Answer 1

Let $f(x)=x+x^2$ and $g(x)=2x^2$. Then $f''(x)=2<4=g''(x)$; however, $f(x)>g(x)$ on $(0,1)$. What matters is that, near $0$, the first derivative of $f$ is larger than the first derivative of $g$.

Answer 2

For differentiable functions,

$$f(x)=g(x)\land\forall y> x: f'(y)<g'(y)\implies \forall y> x:f(y)<g(y)$$

because

$$(g(y)-f(y))'>0$$ so that $g(y)-f(y)$ is a growing function and $g(y)-f(y)>g(x)-f(x)=0$.

You can iterate and write

$$f'(x)=g'(x)\land f(x)=g(x)\land\forall y> x: f''(y)<g''(y)\\\implies f'(x)=g'(x)\land \forall y> x:f'(y)<g'(y)\\\implies \forall y> x:f(y)<g(y).$$