Let $C^1([0,1],\mathbb R)$ be the space of continuously differentiable functions over $[0,1]$. We equip it with the norm $\|f\|_{C^1} = \|f\|_{\infty}+\|f'\|_{\infty}$.
Let $F$ be the linear subspace $F=\{f \in C^1([0,1],\mathbb R) : f(0)=0 \}$ and define the functional $$\varphi: F\to \mathbb R, \quad f\mapsto \int_0^1 \frac{f(t)}t dt$$
What is the norm of $\varphi$ ?
Here are my thoughts: given $f\in F$, since $f(0)=0$ and $f$ is differentiable at $0$, $$f(t)\ln(t) = \frac{f(t)-f(0)}{t-0} \times t\ln(t)\xrightarrow[t\to 0]{} 0 $$
hence integration by parts is licit and yields
$$\varphi(f) = -\int_0^1 f'(t) \ln(t) dt$$
from which we derive $|\varphi(f)| \leq \|f'\|_\infty \int_0^1 |\ln(t)| dt = \|f'\|_\infty\leq \|f\|_{C^1}$.
Therefore $\|\varphi\| \leq 1$. However, this upper bound seems too loose. When experimenting with low-degree polynomials I find that $\displaystyle \frac{|\varphi(f)|}{\|f\|_{C^1}}\leq \frac 12$.
EDIT: Thanks to user Medo I know that $\|\varphi\|>\frac 12$ since for $f:t\mapsto \sin(2t)$, $\varphi(f) \approx 1.60$ while $\|f\|_{C^1} = 1 + 2 = 3$, hence a ratio of $\approx 0.53$.
Best Answer
You can split the integral in the following way to get a better bound. Let $a\in(0,1)$, then
$$ \left\lvert\int_{0}^{1} \frac{f(t)}{t}dt\right\rvert\leq\left\lvert\int_{0}^{a}\frac{f(t)}{t}dt\right\rvert+\left\lvert\int_{a}^{1}\frac{f(t)}{t}dt\right\rvert. $$ Now the second term clearly satisfies$$ \left\lvert\int_{a}^{1}\frac{f(t)}{t}dt\right\rvert\leq \lVert f\rVert_\infty \int_{a}^{1}\frac 1t dt=\lVert f\rVert_\infty (-\ln(a)). $$ The first term on the other hand satisfies, by your own argument with integration by parts $$ \left\lvert\int_{0}^{a} \frac{f(t)}{t}dt\right\rvert\leq \left\lvert\int_{0}^{a}\left(\ln(a)-\ln(t)\right) f'(t)dt\right\rvert\leq a \lVert f'\rVert_\infty. $$ Hence you have $\lvert \varphi(f)\rvert\leq \max(a,-\ln(a))\lVert f \rVert_{C^1}$ for any $a\in(0,1)$. Optimizing in $a$, you find the minimum value of $\min_{a\in(0,1)}\max(a,-\ln(a))\approx 0.567$.